In IB you used real numbers constantly without being told what they are. That was enough for computing. It is not enough for proving, because the big theorems of calculus (the Intermediate Value Theorem, the Mean Value Theorem, the Fundamental Theorem) all rest on one fact: the real number line has no gaps. This module makes that fact precise and shows how much follows from it.

1.1The language of proof

University calculus is written in a compact language of sets and quantifiers. You will read it fluently within a few weeks, but only if you practise it from the start.

The order of quantifiers matters. Compare

The first says “every number has a bigger number”, which is true (take y = x+1). The second says “some number is bigger than every number”, which is false. Same symbols, different order, opposite truth values. The definitions of limit and continuity are long strings of quantifiers, so this is not pedantry.

You already know the main proof techniques from AHL 1.15. In this course you will use all of them:

• Direct proof: assume P, deduce Q.
• Contrapositive: P \Rightarrow Q is equivalent to \neg Q \Rightarrow \neg P.
• Contradiction: assume \neg Q (together with P) and derive something impossible.
• Induction: for statements about every n \in \N.
• Counterexample: one example is enough to disprove a “for all” statement.

1.2From counting numbers to fractions

The natural numbers \N can be added and multiplied but not always subtracted, so we extend to the integers \Z. Integers cannot always be divided, so we extend to the rationals

\Q is an ordered field. That means the familiar rules of algebra hold (addition and multiplication are commutative and associative, multiplication distributes over addition, every number has a negative and every non-zero number has a reciprocal), and so do the rules of inequalities: if a < b then a + c < b + c, and if a<b and c>0 then ac < bc. Every manipulation of equations and inequalities you did in IB follows from these rules.

\Q is also dense: between any two rationals p < q lies another one, \tfrac{p+q}{2}. So the rationals have no “next” number and look as if they fill the line completely. They do not.

1.3A hole in the rationals

Picture the set A = \{q \in \Q : q > 0,\ q^2 < 2\}. Its elements creep upwards: 1,\ 1.4,\ 1.41,\ 1.414,\ \dots They are all below 1.5, so the set is bounded above. But there is no least rational upper bound: if s is a positive rational with s^2 > 2, a slightly smaller rational is still an upper bound, and if s^2 < 2 then s is not an upper bound at all. (The proof of Theorem 1.9 below contains the computation.) The rational line has a hole exactly where \sqrt2 should be. Calculus cannot work on a line with holes: the function x^2 - 2 changes sign on [1,2] without ever being zero in \Q.

1.4Bounds, suprema and infima

A supremum need not belong to the set. If it does, it is the maximum.

• A = [0,1]: \sup A = 1 \in A, so 1 = \max A.
• A = [0,1): \sup A = 1 \notin A. The set has no maximum. To see that 1 is the least upper bound: if 0 \le M < 1 then \tfrac{M+1}{2} is in A and bigger than M, and if M < 0 then 0 \in A beats it. So no number below 1 is an upper bound.
• A = \{\tfrac1n : n \in \N\}: \max A = 1, and \inf A = 0, which is not attained. (That 0 is the greatest lower bound needs the Archimedean property below.)

This is the form in which you will use suprema in practice: just below \sup A there are always elements of A.

1.5The Completeness Axiom

The real numbers \R are an ordered field, with all the algebra and inequality rules of \Q, that satisfies one extra axiom.

\Q satisfies every other axiom and fails this one: the set \{q\in\Q : q>0,\ q^2<2\} has no supremum in \Q. Completeness is the precise meaning of “\R has no holes”. Every time a proof in this course says “let s = \sup\{\dots\}”, this axiom is at work. The matching statement for infima follows from it (Exercise 1.8).

1.6First consequences of completeness

So there are no “infinitely small” positive real numbers. This is why, from Module 3 onwards, we can describe “arbitrarily close” by “closer than every \eps > 0”.

The same idea produces nth roots of every positive number. In Module 5 the Intermediate Value Theorem will give them much more cheaply.

This also explains infinite decimals. The expansion 1.41421356\ldots names the real number \sup\{1,\ 1.4,\ 1.41,\ 1.414,\ \dots\}, and completeness guarantees that this supremum exists. In the same way 0.999\ldots = \sup\{0.9, 0.99, 0.999, \dots\} = 1 (Exercise 1.11).

1.7Absolute value and inequalities

The translation below is used on almost every page of Part I, so make it automatic:

“x is within \delta of a” and “x lies in the open interval of radius \delta centred at a” are the same statement.

Replacing a by a-b and b by b-c gives the distance form |a - c| \le |a - b| + |b - c|: going via b is never shorter. A useful relative is the reverse triangle inequality \big||a| - |b|\big| \le |a-b| (Exercise 1.5).

1.8Beyond IB: how many real numbers are there?

So the rationals, although dense, form a “small” set, and there are two different sizes of infinity on the number line. You will meet this again in measure theory and probability.

Exercises

1. 1.1Fluency Done

   Negate each statement, and decide whether the original is true.

   (a) Every student in the class passed.   (b) \forall x \in \R\ \exists y \in \R : x + y = 0.   (c) \exists M > 0\ \forall x \in \R : |f(x)| \le M, for f(x) = x^2.

   Then decide whether \exists y \in \R\ \forall x \in \R : x + y = 0 is true.

   Solution

   (a) “Some student in the class did not pass.” (b) Negation: \exists x\ \forall y : x + y \ne 0. The original is true: given x, take y = -x. (c) Negation: \forall M > 0\ \exists x : |f(x)| > M. For f(x)=x^2 the negation is true (take x = \sqrt{M}+1), so the original is false.

   The swapped statement is false: whatever y is, the choice x = 1 - y gives x + y = 1 \ne 0. Swapping the quantifiers changed a true statement into a false one.

2. 1.2Fluency Done

   Solve, writing the answer as an interval or union of intervals: (a) |2x - 3| < 5   (b) |x - 1| < |x + 2|   (c) |x^2 - 4| < 1.

   Solution

   (a) -5 < 2x - 3 < 5 \iff -1 < x < 4, so x \in (-1, 4).

   (b) Both sides are non-negative, so we may square: (x-1)^2 < (x+2)^2 \iff -2x + 1 < 4x + 4 \iff x > -\tfrac12. So x \in (-\tfrac12, \infty). Geometrically: x is closer to 1 than to -2.

   (c) -1 < x^2 - 4 < 1 \iff 3 < x^2 < 5 \iff \sqrt3 < |x| < \sqrt5, so x \in (-\sqrt5, -\sqrt3) \cup (\sqrt3, \sqrt5).

3. 1.3Fluency Done

   Find the supremum and infimum of each set, and say whether each is attained (belongs to the set).

   (a) \{\tfrac1n : n \in \N\}   (b) \{x \in \R : x^2 < 3\}   (c) \{(-1)^n(1 - \tfrac1n) : n \in \N\}   (d) \{\tfrac{n}{n+1} : n \in \N\}   (e) \{q \in \Q : q^2 < 2\}, as a subset of \R.

   Solution

   (a) \sup = 1 (attained at n=1), \inf = 0 (not attained; by the Archimedean property no positive number is a lower bound).

   (b) \sup = \sqrt3, \inf = -\sqrt3, neither attained.

   (c) The terms are 0, \tfrac12, -\tfrac23, \tfrac34, -\tfrac45, \dots. \sup = 1 and \inf = -1, neither attained, since every term has absolute value 1 - \tfrac1n < 1 while the even terms approach 1 and the odd terms approach -1.

   (d) \inf = \tfrac12 (attained at n = 1), \sup = 1 (not attained).

   (e) \sup = \sqrt2 and \inf = -\sqrt2, neither attained. As a subset of \Q this set has no supremum at all; inside \R it does. That is completeness at work.

4. 1.4Understanding Done

   Prove that \sqrt3 is irrational. Where exactly does your proof break if you try the same argument for \sqrt4?

   Solution

   Suppose \sqrt3 = p/q with p, q \in \N and no common factor. Then p^2 = 3q^2, so 3 \mid p^2. If 3 \nmid p then p = 3k \pm 1 and p^2 = 9k^2 \pm 6k + 1 leaves remainder 1 on division by 3, which is impossible. So 3 \mid p; write p = 3m. Then 9m^2 = 3q^2, so q^2 = 3m^2 and by the same argument 3 \mid q. This contradicts the assumption that p and q have no common factor.

   For \sqrt4 the step “4 \mid p^2 \Rightarrow 4 \mid p” is false (p = 2 is a counterexample). The argument uses the fact that 3 is prime.

5. 1.5Understanding Done

   Prove the reverse triangle inequality: \big||a| - |b|\big| \le |a - b| for all real a, b.

   Solution

   By the triangle inequality, |a| = |(a - b) + b| \le |a - b| + |b|, so |a| - |b| \le |a-b|. Swapping the roles of a and b gives |b| - |a| \le |b - a| = |a - b|. A number whose value and negative are both at most |a-b| has absolute value at most |a-b|.

6. 1.6Understanding Done

   Let A and B be non-empty sets that are bounded above, and let A + B = \{a + b : a \in A,\ b \in B\}. Prove that \sup(A+B) = \sup A + \sup B.

   Solution

   Let s = \sup A and t = \sup B. For a \in A and b \in B we have a + b \le s + t, so s+t is an upper bound for A+B.

   Now use the ε-test (Lemma 1.6). Let \eps > 0. There are a \in A with a > s - \eps/2 and b \in B with b > t - \eps/2. Then a + b \in A + B and a + b > (s + t) - \eps. Since \eps was arbitrary, s + t = \sup(A+B).

7. 1.7Understanding Done

   Prove Bernoulli’s inequality: if x \ge -1 and n \in \N, then (1+x)^n \ge 1 + nx. (We will use it repeatedly in Module 3.)

   Solution

   Induction on n. For n = 1 both sides equal 1 + x. Suppose (1+x)^n \ge 1 + nx. Since 1 + x \ge 0, multiplying both sides by 1+x preserves the inequality:

   (1+x)^{n+1} \ge (1 + nx)(1+x) = 1 + (n+1)x + nx^2 \ge 1 + (n+1)x.

   The hypothesis x \ge -1 was needed exactly once, to multiply an inequality by 1+x.

8. 1.8Understanding Done

   Show that if A is non-empty and bounded below, then \inf A exists and \inf A = -\sup(-A), where -A = \{-a : a \in A\}.

   Solution

   If m is a lower bound for A, then -m is an upper bound for -A, so by completeness s = \sup(-A) exists. For every a\in A we have -a \le s, so a \ge -s: thus -s is a lower bound for A. If m is any lower bound for A, then -m is an upper bound for -A, so s \le -m, that is m \le -s. Hence -s is the greatest lower bound.

9. 1.9Challenge Done

   Prove that between any two real numbers a < b there is an irrational number.

   Solution

   Apply the density of \Q (Theorem 1.10) to a/\sqrt2 < b/\sqrt2 to get a rational r strictly between them. If r = 0 happens to be chosen, apply density again to the interval (0, b/\sqrt2) or (a/\sqrt2, 0), whichever is non-empty, to get a non-zero rational instead. Then a < r\sqrt2 < b. The number r\sqrt2 is irrational: if r\sqrt2 = q \in \Q then \sqrt2 = q/r \in \Q, contradicting Theorem 1.3.

10. 1.10Challenge Done

    (Nested intervals.) Let [a_1,b_1] \supseteq [a_2,b_2] \supseteq [a_3,b_3] \supseteq \cdots be closed bounded intervals. Prove that some real number lies in all of them. Then show that the corresponding statement for intervals of rationals is false.

    Solution

    Nesting gives a_1 \le a_2 \le \cdots and b_1 \ge b_2 \ge \cdots, with a_n \le b_n. For any n and m, let k = \max(n,m); then a_n \le a_k \le b_k \le b_m. So every b_m is an upper bound for A = \{a_n : n \in \N\}. Let s = \sup A. Then a_n \le s for all n (upper bound) and s \le b_m for all m (least upper bound). So s \in [a_n, b_n] for every n.

    In \Q this fails. Take the decimal truncations of \sqrt2: [1.4, 1.5] \supseteq [1.41, 1.42] \supseteq [1.414, 1.415] \supseteq \cdots (intervals of rationals). The only real number in all of them is \sqrt2, which is not rational, so their intersection in \Q is empty.

11. 1.11Challenge Done

    Prove that 0.999\ldots = 1, where the left side means \sup\{1 - 10^{-n} : n \in \N\}.

    Solution

    Every 1 - 10^{-n} is less than 1, so 1 is an upper bound. Let \eps > 0. By the Archimedean property there is n with 1/n < \eps, and since 10^n > n we get 10^{-n} < 1/n < \eps. Then 1 - 10^{-n} > 1 - \eps. By the ε-test, 1 is the supremum.

12. 1.12Challenge Done

    Show that the set of irrational numbers is uncountable.

    Solution

    Suppose the irrationals could be listed as i_1, i_2, i_3, \dots. The rationals can be listed as q_1, q_2, q_3, \dots. Then q_1, i_1, q_2, i_2, q_3, i_3, \dots would list every real number, contradicting Cantor’s theorem. So “almost all” real numbers are irrational, even though the rationals are dense.

Calculus studies how functions change. Before we can talk about change we need to be exact about what a function is, how functions combine, and when they can be reversed. Much of this module is familiar from IB Topic 2, so read it briskly. Do the exercises anyway: domains, composites and inverses come back on almost every page from Module 6 onwards.

2.1What a function is

The graph of f is the set of points \{(x, f(x)) : x \in A\} in the plane. “Exactly one output for each input” is the vertical line test: no vertical line meets the graph twice. The equation x^2 + y^2 = 1 does not define y as a function of x on [-1,1], because each x in (-1,1) gives two values of y. The formula y = \sqrt{1 - x^2} does define a function: the upper semicircle. Curves like the full circle return in Module 8, where we differentiate them without solving for y.

When a function is given by a formula and no domain is stated, the convention is to take the largest set of real numbers on which the formula makes sense. This is the natural domain. For example

2.2Building new functions

Functions can be added, subtracted, multiplied and divided pointwise: (f+g)(x) = f(x) + g(x), (fg)(x) = f(x)g(x), (f/g)(x) = f(x)/g(x). The domain of each is the intersection of the two domains, with the zeros of g removed for the quotient.

The chain rule in Module 7 differentiates composites layer by layer, so practise seeing the layers. The function h(x) = \sin^3(2x+1) is built in three steps:

The transformations of graphs from IB are compositions with linear functions. The graph of y = a\,f\big(b(x - c)\big) + d is the graph of f stretched horizontally by the factor 1/b, then shifted right by c, stretched vertically by a and shifted up by d.

2.3Injective functions and inverses

An injective function f : A \to \R is a bijection from A onto its range f(A), so it has an inverse defined on f(A). Since (a, b) lies on the graph of f exactly when (b,a) lies on the graph of f^{-1}, the graph of the inverse is the reflection of the graph of f in the line y = x. That reflection is the whole idea behind the derivative of an inverse function in Module 8.

Many functions are not injective, but become injective when we restrict their domain. This is how the inverse trigonometric functions are defined:

2.4Symmetry, periodicity and boundedness

A function is even if f(-x) = f(x) for all x in its domain (graph symmetric in the y-axis), odd if f(-x) = -f(x) (graph symmetric under a half-turn about the origin), and periodic with period p > 0 if f(x + p) = f(x) for all x. For example \cos, |x| and x^2 are even; \sin, \tan and x^3 are odd; e^x is neither. Sine and cosine have period 2\pi; tangent has period \pi. Symmetry saves work: in Module 16 you will see that \int_{-a}^{a} f(x)\,dx = 0 for every odd f.

2.5The catalogue of elementary functions

Calculus at this level is mostly about the following families, and the functions you can build from them with arithmetic and composition.

Angles are always in radians. The point (\cos\theta, \sin\theta) is the point on the unit circle reached by walking an arc of length \theta anticlockwise from (1,0). Radians are not a convention of taste: in Module 4 we will see that \frac{\sin x}{x} \to 1 as x \to 0 only when x is measured in radians, and every derivative of a trigonometric function depends on that limit.

2.6Strange functions for testing ideas

Definitions in calculus are designed for well-behaved functions, but they are understood by testing them on badly behaved ones. Three functions will keep coming back:

• The Dirichlet function D(x) = 1 if x \in \Q and D(x) = 0 if x \notin \Q. Every interval contains both rationals and irrationals (Module 1), so its graph is two dense “clouds” of points. It is discontinuous everywhere (Module 5) and not integrable (Module 14).
• \sin(1/x) for x \ne 0. As x \to 0 it oscillates between -1 and 1 infinitely often, so it has no limit at 0 (Module 4).
• x^2 \sin(1/x), with value 0 at x = 0. It is differentiable everywhere, but its derivative jumps around near 0, so the derivative is not continuous (Modules 6 and 7).

Exercises

1. 2.1Fluency Done

   Find the natural domain and the range: (a) f(x) = \sqrt{4 - x^2}   (b) g(x) = \dfrac{1}{x^2 - 1}   (c) h(x) = \ln(x - 3).

   Solution

   (a) Need 4 - x^2 \ge 0, so the domain is [-2, 2]. As x runs over it, 4 - x^2 takes every value in [0,4], so the range is [0, 2].

   (b) Domain \R \setminus \{-1, 1\}. The quantity x^2 - 1 takes every value in [-1, 0) \cup (0, \infty). Taking reciprocals, [-1,0) goes to (-\infty, -1] and (0,\infty) goes to (0, \infty). Range: (-\infty, -1] \cup (0, \infty).

   (c) Domain (3, \infty); range \R.

2. 2.2Fluency Done

   Let f(x) = \sqrt{x} and g(x) = x^2 - 4. Find f \circ g and g\circ f with their domains, and decide whether g \circ f = f\circ g anywhere.

   Solution

   (f\circ g)(x) = \sqrt{x^2 - 4}, defined when x^2 \ge 4: domain (-\infty, -2] \cup [2, \infty).

   (g\circ f)(x) = (\sqrt{x})^2 - 4 = x - 4, with domain [0, \infty).

   They agree where both are defined and \sqrt{x^2-4} = x - 4. That needs x \ge 4 and x^2 - 4 = x^2 - 8x + 16, so x = \tfrac52, which is not \ge 4. So they agree nowhere.

3. 2.3Fluency Done

   Write each function as a composite of simpler functions, listing the layers from the inside out: (a) \sin^3(2x+1)   (b) e^{\sqrt{x^2+1}}   (c) \ln(\cos x). For (c), find the natural domain.

   Solution

   (a) x \mapsto 2x + 1 \mapsto \sin(\cdot) \mapsto (\cdot)^3.   (b) x \mapsto x^2 + 1 \mapsto \sqrt{\cdot} \mapsto e^{(\cdot)}.   (c) x \mapsto \cos x \mapsto \ln(\cdot).

   For (c) we need \cos x > 0, so the domain is the union of the intervals \big(-\tfrac\pi2 + 2k\pi,\ \tfrac\pi2 + 2k\pi\big), k \in \Z.

4. 2.4Fluency Done

   Find the inverse of f(x) = \dfrac{2x+1}{x-3}, x \ne 3, and state its domain.

   Solution

   Solve y = \frac{2x+1}{x-3} for x: y(x - 3) = 2x + 1, so x(y - 2) = 3y + 1 and x = \frac{3y+1}{y-2}. Hence f^{-1}(x) = \dfrac{3x+1}{x-2}, with domain x \ne 2. (The range of f is \R\setminus\{2\}: y = 2 would need 2x + 1 = 2x - 6.)

5. 2.5Fluency Done

   Evaluate exactly: (a) \arcsin(-\tfrac12)   (b) \arccos(-\tfrac12)   (c) \arctan\sqrt3   (d) \arcsin\big(\sin\tfrac{5\pi}{6}\big)   (e) \cos(\arcsin x) for x \in [-1,1].

   Solution

   (a) -\tfrac\pi6   (b) \tfrac{2\pi}{3}   (c) \tfrac\pi3   (d) \tfrac\pi6, because \tfrac{5\pi}6 \notin [-\tfrac\pi2, \tfrac\pi2] while \sin\tfrac{5\pi}6 = \tfrac12.

   (e) Let \theta = \arcsin x, so \sin\theta = x and \theta \in [-\tfrac\pi2, \tfrac\pi2]. On that interval \cos\theta \ge 0, so \cos\theta = +\sqrt{1 - \sin^2\theta} = \sqrt{1 - x^2}. This identity is exactly what Module 8 needs to differentiate \arcsin.

6. 2.6Understanding Done

   Prove that if f is strictly increasing on its domain, then its inverse f^{-1} is strictly increasing on the range of f.

   Solution

   Let u < v be in the range, with u = f(x) and v = f(y). If x \ge y, then (as f is increasing) f(x) \ge f(y), that is u \ge v, a contradiction. So x < y, i.e. f^{-1}(u) < f^{-1}(v).

7. 2.7Understanding Done

   Show that every function f : \R \to \R is the sum of an even function and an odd function. Identify the two parts when f(x) = e^x.

   Solution

   Write f(x) = \underbrace{\tfrac12\big(f(x) + f(-x)\big)}_{E(x)} + \underbrace{\tfrac12\big(f(x) - f(-x)\big)}_{O(x)}. Replacing x by -x shows E(-x) = E(x) and O(-x) = -O(x).

   For f = e^x the parts are \cosh x = \tfrac12(e^x + e^{-x}) and \sinh x = \tfrac12(e^x - e^{-x}), the hyperbolic cosine and sine. They are part of the NTNU syllabus and reappear in Module 9.

8. 2.8Understanding Done

   Show that \cos(\arctan x) = \dfrac{1}{\sqrt{1+x^2}} and \sin(\arctan x) = \dfrac{x}{\sqrt{1+x^2}} for every real x.

   Solution

   Let \theta = \arctan x \in (-\tfrac\pi2, \tfrac\pi2). There \cos\theta > 0, and 1 + \tan^2\theta = \sec^2\theta gives \cos^2\theta = \frac{1}{1+x^2}, so \cos\theta = \frac{1}{\sqrt{1+x^2}}. Then \sin\theta = \tan\theta\cos\theta = \frac{x}{\sqrt{1+x^2}}.

9. 2.9Understanding Done

   Show that f(x) = \dfrac{x}{1 + |x|} is a bijection from \R onto (-1, 1), and find f^{-1}.

   Solution

   For x \ge 0, f(x) = 1 - \frac{1}{1+x}, which is strictly increasing, takes the value 0 at x=0, and takes every value in [0,1) (solve y = \frac{x}{1+x} to get x = \frac{y}{1-y} \ge 0). Since f is odd, it maps (-\infty, 0] onto (-1, 0] in the same way. So f is strictly increasing on \R, hence injective, with range (-1,1).

   The inverse is f^{-1}(y) = \dfrac{y}{1 - |y|} for -1 < y < 1. So \R and the bounded interval (-1,1) are “the same size” in the sense of Module 1.

10. 2.10Challenge Done

    Suppose f : \R \to \R satisfies f(x + y) = f(x) + f(y) for all x, y. Prove that f(0) = 0, that f(-x) = -f(x), and that f(q) = q\,f(1) for every rational q. (Module 5 finishes the job: if f is also continuous, then f(x) = cx for all real x.)

    Solution

    f(0) = f(0 + 0) = 2f(0), so f(0) = 0. Then 0 = f(x + (-x)) = f(x) + f(-x). By induction, f(nx) = n f(x) for n \in \N: the step is f((n+1)x) = f(nx) + f(x). Together with oddness, f(mx) = m f(x) for all m \in \Z.

    Now let q = m/n with m\in\Z, n \in \N. Then n\,f(q) = f(nq) = f(m) = m f(1), so f(q) = \frac mn f(1) = q f(1).

11. 2.11Challenge Done

    Show that every positive rational number is a period of the Dirichlet function, that no irrational number is a period, and hence that the Dirichlet function has no smallest positive period.

    Solution

    If p \in \Q, then x + p is rational exactly when x is, so D(x + p) = D(x) for every x. If p is irrational then D(0 + p) = 0 \ne 1 = D(0), so p is not a period. The periods are therefore exactly the positive rationals, and there is no smallest positive rational (if p is one, so is p/2).

All of calculus is built on one idea, the limit. Derivatives are limits of difference quotients, integrals are limits of sums, and series are limits of partial sums. The simplest setting for limits is a sequence: a list of numbers a_1, a_2, a_3, \dots This module defines convergence precisely and proves the results you will use again and again. It covers the “understanding of limits (convergence and divergence)” of AHL 5.12 in far more depth than the IB requires, and it is the core of the first weeks at NTNU.

3.1Sequences

A sequence is a function from \N to \R, written n \mapsto a_n or simply (a_n). følge Some sequences that will matter:

• a_n = \frac1n: \ 1, \frac12, \frac13, \frac14, \dots
• a_n = (-1)^n: \ -1, 1, -1, 1, \dots
• a_n = \left(1 + \frac1n\right)^n: \ 2,\ 2.25,\ 2.370\ldots,\ 2.441\ldots,\ \dots
• a recursive sequence: a_1 = 1, a_{n+1} = \sqrt{2 + a_n}: \ 1,\ 1.732\ldots,\ 1.931\ldots,\ 1.982\ldots,\ \dots

A table like this suggests a limit. It cannot prove one: the values could start drifting at n = 10^{12}. To prove anything we need a definition.

3.2What convergence means

Read it as a game. Your opponent names a tolerance \eps, however small. You must reply with a point N in the sequence beyond which every term stays within \eps of L. If you can always reply, the sequence converges to L. Usually N depends on \eps: a smaller tolerance needs a later starting point.

This is the pattern for every convergence proof: find N by working backwards from the inequality you want, then write the argument forwards.

3.3Rules for limits

Strict inequalities are not preserved: 0 < \frac1n for every n, yet both sides have limit 0.

For example \frac{\sin n}{n} \to 0, because -\frac1n \le \frac{\sin n}{n} \le \frac1n and both bounds tend to 0.

3.4Monotone sequences and completeness

To use the definition of convergence you must already know the limit. The next theorem proves convergence without knowing the limit in advance, and it is the first place where completeness pays off.

In \Q this theorem fails: 1, 1.4, 1.41, 1.414, \dots is increasing and bounded but has no rational limit.

In Module 9 this same number turns out to be the base whose exponential function is its own derivative, and in Exercise 3.12 you prove e = \sum_{k=0}^\infty \frac{1}{k!}.

3.5A toolkit of standard limits

For large n these give a hierarchy that is worth memorising: \ \ln n \ll n^k \ll b^n \ll n! \ll n^n. Module 11 proves the corresponding facts for functions with L’Hôpital’s rule.

3.6Subsequences and the Bolzano–Weierstrass Theorem

A subsequence of (a_n) is a sequence a_{n_1}, a_{n_2}, a_{n_3}, \dots with n_1 < n_2 < n_3 < \cdots. If a_n \to L then every subsequence also tends to L, since n_k \ge k. So if two subsequences have different limits, the sequence diverges: the even terms of (-1)^n tend to 1 and the odd terms to -1.

This theorem is the engine inside the Extreme Value Theorem in Module 5.

3.7Series

An infinite sum is not a sum of infinitely many numbers (nobody can do that). It is the limit of finite sums, and so everything about series is a statement about the sequence (S_n).

The partial sums grow very slowly, roughly like \ln n: after a million terms the sum is only about 14.4.

Telescoping series can be summed exactly. Since \frac{1}{k(k+1)} = \frac1k - \frac1{k+1},

Exercises

1. 3.1Fluency Done

   Find each limit, justifying the steps with the limit laws or the Squeeze Theorem.

   (a) \dfrac{n^3 - 2n}{4n^3 + n^2 + 1}   (b) \sqrt{n^2 + n} - n   (c) \dfrac{3^n + 2^n}{3^{n+1} - 2^n}   (d) \dfrac{\cos n}{\sqrt n}

   Solution

   (a) Divide by n^3: \frac{1 - 2/n^2}{4 + 1/n + 1/n^3} \to \frac14.

   (b) Multiply by the conjugate: \sqrt{n^2+n} - n = \frac{(n^2 + n) - n^2}{\sqrt{n^2+n} + n} = \frac{n}{\sqrt{n^2+n}+n} = \frac{1}{\sqrt{1 + 1/n} + 1} \to \frac12.

   (c) Divide by 3^n: \frac{1 + (2/3)^n}{3 - (2/3)^n} \to \frac{1 + 0}{3 - 0} = \frac13, using (2/3)^n \to 0.

   (d) -\frac{1}{\sqrt n} \le \frac{\cos n}{\sqrt n} \le \frac1{\sqrt n}, and \frac{1}{\sqrt n} \to 0, so the limit is 0 by squeezing.

2. 3.2Fluency Done

   Find the sum of each series, or explain why it diverges.

   (a) \displaystyle\sum_{k=1}^\infty 3\left(-\tfrac12\right)^k   (b) \displaystyle\sum_{k=0}^\infty \frac{2^k + 3^k}{6^k}   (c) 0.272727\ldots written as a fraction   (d) \displaystyle\sum_{k=0}^\infty (x - 1)^k for the values of x where it converges.

   Solution

   (a) A geometric series with first term -\frac32 and ratio -\frac12: the sum is \frac{-3/2}{1 + 1/2} = -1.

   (b) \sum \left(\frac13\right)^k + \sum\left(\frac12\right)^k = \frac{3}{2} + 2 = \frac72 (the sum of two convergent series is the limit of the sum of their partial sums).

   (c) 0.2727\ldots = 27\left(\frac{1}{100} + \frac{1}{100^2} + \cdots\right) = 27\cdot\frac{1/100}{1 - 1/100} = \frac{27}{99} = \frac{3}{11}.

   (d) Converges exactly when |x - 1| < 1, i.e. 0 < x < 2, with sum \frac{1}{1 - (x-1)} = \frac{1}{2 - x}.

3. 3.3Fluency Done

   Find \displaystyle\sum_{k=1}^\infty \frac{1}{k^2 + 3k + 2}.

   Solution

   k^2 + 3k + 2 = (k+1)(k+2) and \frac{1}{(k+1)(k+2)} = \frac1{k+1} - \frac1{k+2}. The partial sums telescope: S_n = \frac12 - \frac{1}{n+2} \to \frac12.

4. 3.4Understanding Done

   Prove directly from the definition that \dfrac{2n+3}{n+1} \to 2.

   Solution

   \left|\frac{2n+3}{n+1} - 2\right| = \frac{1}{n+1} < \frac1n. Given \eps > 0, choose N > 1/\eps. For n \ge N, \left|\frac{2n+3}{n+1} - 2\right| < \frac1n \le \frac1N < \eps.

5. 3.5Understanding Done

   Prove that if a_n \to L then |a_n| \to |L|. Is the converse true? Is it true when L = 0?

   Solution

   By the reverse triangle inequality, \big||a_n| - |L|\big| \le |a_n - L|. So any N that works for (a_n) and \eps also works for (|a_n|).

   The converse is false: |(-1)^n| \to 1 but (-1)^n diverges. When L = 0 it is true, because |a_n - 0| = \big||a_n| - 0\big|, so the two statements are literally the same.

6. 3.6Understanding Done

   Suppose b_n \to b with b \ne 0 and every b_n \ne 0. Show that |b_n| > \frac{|b|}{2} for all large n, and deduce that \frac{1}{b_n} \to \frac1b. Combine with the product law to obtain the quotient law.

   Solution

   Take \eps = \frac{|b|}{2}: for n \ge N_1, |b_n - b| < \frac{|b|}2, so |b_n| \ge |b| - |b - b_n| > \frac{|b|}{2}.

   Then for n \ge N_1, \ \left|\frac1{b_n} - \frac1b\right| = \frac{|b - b_n|}{|b_n|\,|b|} < \frac{2}{|b|^2}|b_n - b|. Given \eps>0, choose N \ge N_1 with |b_n - b| < \frac{\eps|b|^2}{2} for n \ge N; then \left|\frac{1}{b_n} - \frac1b\right| < \eps.

   Finally \frac{a_n}{b_n} = a_n\cdot\frac1{b_n} \to a\cdot\frac1b by the product law.

7. 3.7Understanding Done

   (The Babylonian method.) Let a_1 = 2 and a_{n+1} = \frac12\left(a_n + \frac{3}{a_n}\right). Show that a_n \ge \sqrt3 for all n \ge 2, that (a_n) is decreasing, and that a_n \to \sqrt3. Compute a_2, a_3, a_4.

   Solution

   All terms are positive. By AM–GM, a_{n+1} = \frac12\left(a_n + \frac3{a_n}\right) \ge \sqrt{a_n\cdot\frac{3}{a_n}} = \sqrt3. (Also a_1 = 2 > \sqrt3.)

   Decreasing: a_{n+1} - a_n = \frac{3 - a_n^2}{2a_n} \le 0 since a_n^2 \ge 3.

   By the Monotone Convergence Theorem a_n \to L \ge \sqrt 3 > 0. Taking limits in the recursion (quotient law, L \ne 0): L = \frac12(L + \frac3L), so L^2 = 3 and L = \sqrt3.

   a_2 = 1.75, a_3 = 1.732142\ldots, a_4 = 1.7320508\ldots, against \sqrt3 = 1.7320508\ldots. The number of correct digits roughly doubles each step; Module 13 explains why (this is Newton’s method).

8. 3.8Understanding Done

   Decide whether each series converges: (a) \sum \frac{1}{\sqrt k}   (b) \sum \frac{1}{k^2 + k + 1}   (c) \sum \frac{k}{2^k} (use the ratio test)   (d) \sum \frac{k!}{k^k}.

   Solution

   (a) Diverges: \frac{1}{\sqrt k} \ge \frac1k and the harmonic series diverges.

   (b) Converges: 0 < \frac1{k^2+k+1} < \frac1{k^2}.

   (c) Converges: \frac{a_{k+1}}{a_k} = \frac{k+1}{2k} \to \frac12 < 1. (Its sum is 2; see Module 19.)

   (d) Converges: \frac{a_{k+1}}{a_k} = \frac{(k+1)!\,k^k}{(k+1)^{k+1}\,k!} = \left(\frac{k}{k+1}\right)^k = \frac{1}{(1 + 1/k)^k} \to \frac1e < 1.

9. 3.9Fluency Done

   Use subsequences to show that a_n = (-1)^n + \frac1n diverges.

   Solution

   The even terms 1 + \frac{1}{2m} \to 1 and the odd terms -1 + \frac{1}{2m-1} \to -1. If (a_n) converged, every subsequence would have the same limit.

10. 3.10Challenge Done

    Prove that a^{1/n} \to 1 for every a > 0. (Hint: for a > 1 write a^{1/n} = 1 + h_n and use Bernoulli’s inequality.)

    Solution

    For a > 1: a^{1/n} = 1 + h_n with h_n > 0. Bernoulli gives a = (1 + h_n)^n \ge 1 + nh_n, so 0 < h_n \le \frac{a - 1}{n} \to 0. For a = 1 it is trivial. For 0 < a < 1, a^{1/n} = \frac{1}{(1/a)^{1/n}} \to \frac11 = 1 by the first case and the quotient law.

11. 3.11Challenge Done

    Prove that n^{1/n} \to 1. (Hint: write n^{1/n} = 1 + h_n and keep only the \binom n2 term of the binomial expansion.)

    Solution

    With h_n \ge 0, for n \ge 2: \ n = (1 + h_n)^n \ge \binom{n}{2}h_n^2 = \frac{n(n-1)}{2}h_n^2. So h_n^2 \le \frac{2}{n-1} and 0 \le h_n \le \sqrt{\frac{2}{n-1}} \to 0.

12. 3.12Challenge Done

    Prove that e = \displaystyle\sum_{k=0}^\infty \frac{1}{k!}. (Hint: the proof of Theorem 3.14 shows e_n \le s_n := \sum_{k=0}^n \frac1{k!}. For the other direction fix m, keep only the terms k \le m in the expansion of e_n for n \ge m, and let n \to \infty.)

    Solution

    The series has positive terms and its partial sums s_n are bounded by 3 (same estimate as in the theorem), so s = \sum_{k=0}^\infty \frac1{k!} exists, and e_n \le s_n \le s gives e \le s.

    Fix m. For n \ge m, dropping the terms with k > m,

    e_n \ge \sum_{k=0}^{m}\frac1{k!}\left(1 - \tfrac1n\right)\cdots\left(1 - \tfrac{k-1}n\right).

    This is a finite sum of terms each tending to \frac1{k!} as n \to \infty, so letting n\to\infty (limits preserve \le) gives e \ge s_m. Now let m \to \infty: e \ge s. Hence e = s.

13. 3.13Challenge Done

    Prove that e is irrational. (Hint: if e = p/q, consider x = q!\left(e - \sum_{k=0}^{q}\frac1{k!}\right).)

    Solution

    Suppose e = \frac pq with p,q \in \N. Then x = q!\,e - \sum_{k=0}^q \frac{q!}{k!} is an integer, since q!\,e = (q-1)!\,p and each \frac{q!}{k!} is an integer. By the previous exercise x = \sum_{k=q+1}^\infty \frac{q!}{k!} > 0. But

    x = \frac{1}{q+1} + \frac{1}{(q+1)(q+2)} + \cdots < \frac1{q+1} + \frac1{(q+1)^2} + \cdots = \frac{1}{q} \le 1.

    So x is an integer strictly between 0 and 1, which is impossible.

A sequence approaches its limit as n \to \infty. A function can approach a limit as x approaches any point a, from both sides at once, and this is the kind of limit derivatives need: the difference quotient \frac{f(a+h) - f(a)}{h} is not even defined at h = 0. This module defines \lim_{x\to a} f(x), shows how to compute and prove limits, and proves the trigonometric limit that every derivative of \sin and \cos depends on.

4.1Approaching without arriving

The function f(x) = \frac{x^2 - 1}{x - 1} is undefined at x = 1. Everywhere else it equals x + 1, so as x gets close to 1 its values get close to 2. A more interesting case is \frac{\sin x}{x}, which is also undefined at 0:

And a physical example: a stone dropped from rest falls s(t) = 4.9t^2 metres in t seconds. Its average velocity between t = 1 and t = 1 + h is

The instantaneous velocity at t = 1 should be what this approaches as h \to 0, namely 9.8 m/s. We need a definition of “approaches” that never asks for the value at the point itself.

4.2The ε–δ definition

Three remarks.

• The condition 0 < |x - a| excludes x = a. The value f(a) plays no part: it may be undefined, or different from L.
• It is the same game as for sequences. Your opponent names \eps (a horizontal band L \pm \eps); you must find \delta (a vertical window a \pm \delta) so that the graph over the window, apart from the point a itself, stays inside the band.
• If some \delta works, every smaller \delta works too. So you never need the best \delta, only one that works.

4.3Limits through sequences

The next theorem connects function limits to the sequence limits of Module 3. It means we get all the limit laws for functions for free.

Since \lim_{x\to a} x = a and constants have constant limits, it follows that for every polynomial P, \lim_{x\to a}P(x) = P(a), and for rational functions \lim_{x\to a}\frac{P(x)}{Q(x)} = \frac{P(a)}{Q(a)} whenever Q(a) \ne 0. Direct substitution works for polynomials and rational functions away from zeros of the denominator. At a point where substitution gives \frac00, simplify first: factorise and cancel, multiply by a conjugate, or combine fractions.

The sequential characterisation is also the best tool for showing that a limit does not exist: find two sequences approaching a along which f has different limits.

4.4One-sided limits

For the sign function \operatorname{sgn} x = \frac{x}{|x|} the right-hand limit at 0 is 1 and the left-hand limit is -1, so there is no limit. For \sqrt x at 0 only the right-hand limit makes sense (=0), because the function is not defined to the left.

4.5Limits involving infinity

These describe asymptotes. The line y = L is a horizontal asymptote if f(x) \to L as x \to \infty or x \to -\infty. The line x = a is a vertical asymptote if f(x) \to \pm\infty as x \to a from at least one side. The line y = mx + c is an oblique asymptote if f(x) - (mx + c) \to 0 as x \to \pm\infty. asymptote

For a rational function as x \to \pm\infty, divide numerator and denominator by the highest power of x in the denominator:

A useful trick: the substitution t = 1/x turns a limit as x \to \infty into a limit as t \to 0^+.

4.6The Squeeze Theorem and the key trigonometric limits

Exercises

1. 4.1Fluency Done

   Evaluate: (a) \displaystyle\lim_{x\to3}\frac{x^2 - 9}{x^2 - 2x - 3}   (b) \displaystyle\lim_{x\to0}\frac{\sqrt{1+x} - 1}{x}   (c) \displaystyle\lim_{x\to0}\frac{\sin 7x}{3x}   (d) \displaystyle\lim_{x\to0}\frac{\tan 3x}{\sin 2x}   (e) \displaystyle\lim_{x\to\infty}\frac{2x^3 - x}{4x^3 + x^2 + 1}   (f) \displaystyle\lim_{x\to\infty}\left(\sqrt{x^2 + 4x} - x\right)

   Solution

   (a) \frac{(x-3)(x+3)}{(x-3)(x+1)} = \frac{x+3}{x+1} \to \frac64 = \frac32.

   (b) \frac{(1+x) - 1}{x(\sqrt{1+x}+1)} = \frac{1}{\sqrt{1+x}+1}\to\frac12.

   (c) \frac73\cdot\frac{\sin7x}{7x} \to \frac73.

   (d) \frac{\tan 3x}{\sin 2x} = \frac{\sin 3x}{3x}\cdot\frac{2x}{\sin 2x}\cdot\frac{1}{\cos 3x}\cdot\frac32 \to 1\cdot1\cdot1\cdot\frac32 = \frac32.

   (e) Divide by x^3: \frac{2 - 1/x^2}{4 + 1/x + 1/x^3} \to \frac12.

   (f) \frac{(x^2+4x) - x^2}{\sqrt{x^2+4x}+x} = \frac{4x}{\sqrt{x^2+4x} + x} = \frac{4}{\sqrt{1+4/x}+1} \to 2 (for x>0, \sqrt{x^2} = x).

2. 4.2Fluency Done

   Let f(x) = x^2 + 1 for x < 1 and f(x) = ax + 3 for x \ge 1. Find the value of a for which \lim_{x\to1} f(x) exists.

   Solution

   The left limit is 2 and the right limit is a + 3. They agree when a = -1.

3. 4.3Fluency Done

   Find all asymptotes of y = \dfrac{2x^2 + 3x - 1}{x - 1}.

   Solution

   Vertical: x = 1 (the numerator is 4 \ne 0 there, so y \to \pm\infty). Long division: 2x^2 + 3x - 1 = (x-1)(2x+5) + 4, so y = 2x + 5 + \frac{4}{x-1}. Since \frac{4}{x-1} \to 0 as x \to\pm\infty, the oblique asymptote is y = 2x + 5. There is no horizontal asymptote.

4. 4.4Understanding Done

   Prove from the ε–δ definition: (a) \lim_{x\to-1}(4x + 7) = 3   (b) \lim_{x\to2}x^2 = 4   (c) \lim_{x\to1}\frac1x = 1.

   Solution

   (a) |4x + 7 - 3| = 4|x + 1|; take \delta = \eps/4.

   (b) |x^2 - 4| = |x - 2||x + 2|. If |x - 2| < 1 then 1 < x < 3 and |x + 2| < 5. Take \delta = \min(1, \eps/5); then |x^2 - 4| < 5\cdot\frac\eps5 = \eps.

   (c) \left|\frac1x - 1\right| = \frac{|x - 1|}{|x|}. If |x - 1| < \frac12 then x > \frac12 and \frac{1}{|x|} < 2. Take \delta = \min\left(\frac12, \frac\eps2\right); then \left|\frac1x - 1\right| < 2\cdot\frac\eps2 = \eps.

5. 4.5Understanding Done

   Show that \lim_{x\to0}x^2\sin(1/x) = 0, and explain why this does not contradict the fact that \lim_{x\to0}\sin(1/x) does not exist.

   Solution

   -x^2 \le x^2\sin(1/x) \le x^2 for x \ne 0, and \pm x^2 \to 0, so the limit is 0 by the Squeeze Theorem. The product law needs both factors to have limits. Here \sin(1/x) has none, but it is bounded, and a bounded factor times something tending to 0 tends to 0.

6. 4.6Understanding Done

   Write out the negation of “\lim_{x\to a} f(x) = L”. Use it to prove that \lim_{x\to0}\operatorname{sgn} x does not exist, by showing that no number L is the limit.

   Solution

   Negation: there is an \eps > 0 such that for every \delta > 0 there is an x with 0 < |x - a| < \delta and |f(x) - L| \ge \eps.

   Let L be any number and take \eps = 1. For any \delta > 0, the points x = \pm\frac\delta2 give \operatorname{sgn} x = \pm1. Since |1 - L| + |L + 1| \ge 2, at least one of |1 - L| and |-1 - L| is \ge 1, so one of the two points satisfies |\operatorname{sgn} x - L| \ge 1. So no L is the limit.

7. 4.7Understanding Done

   Show that \displaystyle\lim_{x\to0}\frac{1 - \cos x}{x^2} = \frac12.

   Solution

   \frac{1 - \cos x}{x^2} = \frac{1 - \cos^2 x}{x^2(1 + \cos x)} = \left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{1 + \cos x} \to 1\cdot\frac12 = \frac12. So for small x, \cos x \approx 1 - \frac{x^2}{2}. Module 19 will explain this as the start of a Maclaurin series.

8. 4.8Understanding Done

   Prove that if \lim_{x\to a} f(x) = L > 0, then f(x) > \frac L2 > 0 for all x \ne a in some open interval around a. (We will use this repeatedly: a function with a positive limit is positive nearby.)

   Solution

   Take \eps = \frac L2 in the definition: there is \delta > 0 with |f(x) - L| < \frac L2 when 0 < |x - a| < \delta. Then f(x) > L - \frac L2 = \frac L2. The same argument with |f(x)| \ge |L| - |f(x) - L| shows that a non-zero limit keeps f away from 0 near a, which the quotient law needs.

9. 4.9Understanding Done

   Show that \lim_{x\to0} x\left\lfloor \frac1x\right\rfloor = 1, where \lfloor y\rfloor is the greatest integer \le y.

   Solution

   For every y, y - 1 < \lfloor y \rfloor \le y. With y = \frac1x: if x > 0, multiplying by x gives 1 - x < x\lfloor 1/x\rfloor \le 1. If x < 0 the inequalities reverse: 1 \le x\lfloor1/x\rfloor < 1 - x. Either way x\lfloor 1/x\rfloor lies between 1 and 1 - x, both of which tend to 1, so the limit is 1 by squeezing.

10. 4.10Challenge Done

    Let f(x) = x if x is rational and f(x) = 0 if x is irrational. Show that \lim_{x\to0}f(x) = 0, but that \lim_{x\to a} f(x) does not exist for any a \ne 0.

    Solution

    |f(x)| \le |x| for all x, so \delta = \eps works at 0.

    Let a \ne 0. By density (Theorem 1.10 and Exercise 1.9) there are rationals q_n \to a and irrationals r_n \to a, all different from a. Then f(q_n) = q_n \to a and f(r_n) = 0 \to 0. These limits differ, so by Theorem 4.6 there is no limit at a.

11. 4.11Challenge Done

    Prove the product law for function limits directly from the ε–δ definition, without using sequences.

    Solution

    Suppose f \to L and g \to M as x \to a. Write |f(x)g(x) - LM| \le |f(x)|\,|g(x) - M| + |M|\,|f(x) - L|.

    Choose \delta_1 so that |f(x) - L| < 1 for 0<|x-a|<\delta_1; then |f(x)| < |L| + 1. Given \eps > 0, choose \delta_2 with |g(x) - M| < \frac{\eps}{2(|L|+1)} and \delta_3 with |f(x) - L| < \frac{\eps}{2(|M|+1)}. For 0 < |x - a| < \min(\delta_1,\delta_2,\delta_3),

    |f(x)g(x) - LM| < (|L|+1)\frac{\eps}{2(|L|+1)} + |M|\frac{\eps}{2(|M|+1)} < \frac\eps2 + \frac\eps2 = \eps.
12. 4.12Challenge Done

    Prove that if \lim_{x\to a} f(x) = L then \lim_{x\to a}|f(x)| = |L|. Show by example that the converse fails, and prove that it holds when L = 0.

    Solution

    \big||f(x)| - |L|\big| \le |f(x) - L|, so the same \delta works. The converse fails for \operatorname{sgn} x at 0: |\operatorname{sgn} x| = 1 for x \ne 0, so its limit is 1, but \operatorname{sgn} x has no limit. When L = 0, the statements “|f(x) - 0| < \eps” and “\big||f(x)| - 0\big| < \eps” are identical.

Continuity is the precise version of “drawing the graph without lifting the pen”. A continuous function on a closed interval has two guaranteed behaviours: it takes every value in between (the Intermediate Value Theorem), and it attains a largest and a smallest value (the Extreme Value Theorem). Both fail over \Q, and both proofs run through completeness. Every optimisation problem you solved in IB relied silently on the second one.

5.1Continuity at a point

Three things must happen: f(a) exists, \lim_{x\to a} f(x) exists, and the two are equal. (The restriction 0 < |x - a| from the definition of a limit can be dropped, since at x = a the inequality holds trivially.) Combining with Theorem 4.6: f is continuous at a exactly when x_n \to a implies f(x_n) \to f(a). In words, continuous functions are the ones you can take limits inside: f(\lim x_n) = \lim f(x_n).

When continuity fails, it fails in one of a few recognisable ways:

The informal IB description (AHL 5.12) is the graph-has-no-breaks picture. The definition above is what that picture means, and it covers functions whose graphs you could never draw, such as the Dirichlet function, which is continuous nowhere (Exercise 5.7).

5.2Building continuous functions

The same argument shows that limits pass through continuous functions: if \lim_{x\to a}g(x) = b and f is continuous at b, then \lim_{x\to a} f(g(x)) = f(b). For example \lim_{x\to0}\cos\left(\frac{\sin x}{x}\right) = \cos 1.

Combining these results with Theorem 5.8 below and Module 9, every elementary function is continuous on its natural domain: polynomials, rational functions, roots, |x|, trigonometric and inverse trigonometric functions, exponentials and logarithms, and every function built from them by arithmetic and composition. This is why direct substitution gives the limit of such a function at any point of its domain.

5.3The Intermediate Value Theorem

Both hypotheses matter. Over \Q the theorem fails: x^2 - 2 is negative at 1 and positive at 2 but never zero at a rational point. Without continuity it fails too: \operatorname{sgn} x takes the values -1 and 1 on [-1,1] but never \frac12.

The IVT is also an algorithm. If f(a) and f(b) have opposite signs, evaluate f at the midpoint m = \frac{a+b}2 and keep whichever half still has a sign change. After n halvings the root is trapped in an interval of length \frac{b-a}{2^n}. This is the bisection method. Run forever, it gives a second proof of the IVT (for k = 0, which is enough after subtracting k): the halved intervals are nested, so by Exercise 1.10 some point c lies in all of them, and their endpoints converge to c from both sides. By continuity f(c) is a limit of values \le 0 and also of values \ge 0, so f(c) = 0.

def bisect(f, a, b, tol=1e-10):
    # assumes f(a) and f(b) have opposite signs
    while b - a > tol:
        m = (a + b) / 2
        if f(a) * f(m) <= 0:
            b = m
        else:
            a = m
    return (a + b) / 2

print(bisect(lambda x: x**3 + x - 1, 0, 1))   # 0.6823278038...

You will write programs like this in your first semester at NTNU, where numerical methods are taught alongside the theory.

5.4The Extreme Value Theorem

Every hypothesis is needed:

• \frac1x is continuous on (0, 1] but unbounded: the interval is not closed.
• x is continuous on (0, 1) and bounded, but has no maximum or minimum.
• x on [0, \infty) has no maximum: the interval is not bounded.
• f(x) = x for 0 \le x < 1 with f(1) = 0 has supremum 1 on [0,1], never attained: f is not continuous.

The EVT is what justifies the method you used for optimisation in IB. A continuous function on [a,b] has a maximum; Module 10 shows that at an interior maximum the derivative is zero; so the maximum must be at an endpoint or at a critical point, and comparing those finitely many values finds it.

5.5Continuity of inverse functions

So x^{1/n} (the inverse of x^n on [0,\infty)), \arcsin, \arccos, \arctan, and in Module 9 the logarithm, are all continuous.

Exercises

1. 5.1Fluency Done

   Classify the discontinuity: (a) \frac{x^2-4}{x-2} at 2   (b) \lfloor x\rfloor at 3   (c) \frac{1}{(x-1)^2} at 1   (d) \sin\frac1x at 0. For any removable discontinuity, say how to redefine the function to remove it.

   Solution

   (a) Removable: the limit is 4; define f(2) = 4, giving the function x + 2. (b) Jump: left limit 2, right limit 3. (c) Infinite. (d) Oscillating: no one-sided limits (Example 4.9).

2. 5.2Fluency Done

   Find k so that f(x) = kx^2 + 1 for x \le 2 and f(x) = 3x - k for x > 2 is continuous everywhere.

   Solution

   Both pieces are continuous, so only x = 2 matters. The left value and limit is 4k + 1; the right limit is 6 - k. Setting 4k + 1 = 6 - k gives k = 1.

3. 5.3Fluency Done

   Show that x^5 - 3x + 1 = 0 has at least three real roots.

   Solution

   Let f(x) = x^5 - 3x + 1, a continuous function. f(-2) = -25 < 0, f(0) = 1 > 0, f(1) = -1 < 0, f(2) = 27 > 0. By the IVT there is a root in each of (-2,0), (0,1) and (1,2), and these intervals are disjoint.

4. 5.4Fluency Done

   Use four steps of bisection on [0, 1] to locate the root of x^3 + x - 1 = 0 in an interval of length \frac1{16}.

   Solution

   f(0) = -1, f(1) = 1. Midpoint 0.5: f = -0.375, keep [0.5, 1]. Midpoint 0.75: f \approx 0.172, keep [0.5, 0.75]. Midpoint 0.625: f \approx -0.131, keep [0.625, 0.75]. Midpoint 0.6875: f \approx 0.0125, keep [0.625, 0.6875]. The root (about 0.6823) lies in [0.625, 0.6875].

5. 5.5Understanding Done

   Prove from the ε–δ definition that f(x) = x^2 is continuous at every real number a.

   Solution

   |x^2 - a^2| = |x - a|\,|x + a|. If |x - a| < 1 then |x + a| \le |x - a| + 2|a| < 1 + 2|a|. Given \eps > 0, let \delta = \min\left(1, \frac{\eps}{1 + 2|a|}\right). If |x - a| < \delta then |x^2 - a^2| < (1 + 2|a|)\cdot\frac{\eps}{1 + 2|a|} = \eps. Note that \delta depends on a as well as \eps: x^2 is not uniformly continuous on \R.

6. 5.6Understanding Done

   (A fixed-point theorem.) Prove that if f : [0,1] \to [0,1] is continuous, then f(c) = c for some c \in [0,1].

   Solution

   Let g(x) = f(x) - x, continuous on [0,1]. Since f takes values in [0,1], g(0) = f(0) \ge 0 and g(1) = f(1) - 1 \le 0. By the IVT, g(c) = 0 for some c \in[0,1], that is f(c) = c. (Brouwer’s theorem generalises this to higher dimensions.)

7. 5.7Understanding Done

   Show that the Dirichlet function D is discontinuous at every point, and that g(x) = x\,D(x) is continuous at 0 and nowhere else.

   Solution

   Let a \in \R. There are rationals q_n \to a and irrationals r_n \to a (density). Then D(q_n) = 1 and D(r_n) = 0, so the two sequences of values have different limits, and \lim_{x\to a} D(x) does not exist. So D is not continuous at a.

   For g: |g(x)| \le |x|, so g(x) \to 0 = g(0) as x \to 0. At a \ne 0 the same two sequences give g(q_n) = q_n \to a and g(r_n) = 0, and a \ne 0, so there is no limit.

8. 5.8Understanding Done

   Give an example of each, and say which hypothesis of the EVT fails: (a) a continuous function on (0,1) with no maximum; (b) a continuous function on [0,\infty) that is unbounded; (c) a function on [0,1] with no maximum.

   Solution

   (a) f(x) = x: the interval is not closed. (b) f(x) = x: the interval is not bounded. (c) f(x) = x for x < 1, f(1) = 0: the function is not continuous at 1. (Many other examples work.)

9. 5.9Challenge Done

   Continue Exercise 2.10: if f is continuous and f(x + y) = f(x) + f(y) for all x, y, prove that f(x) = cx for every real x, where c = f(1).

   Solution

   We know f(q) = cq for every rational q. Given a real x, choose rationals q_n \to x (density of \Q). By continuity f(x) = \lim f(q_n) = \lim cq_n = cx.

   (Without continuity there are wild solutions that are not of this form, but constructing them needs the axiom of choice.)

10. 5.10Challenge Done

    A hiker starts up a mountain path at 08:00 and reaches the top at 16:00. The next day she starts down the same path at 08:00 and reaches the bottom at 16:00, walking at a completely different and varying pace. Prove that there is a time of day at which she is at exactly the same point of the path on both days.

    Solution

    Measure position by the distance u(t) (day 1) and d(t) (day 2) along the path from the bottom, at time t \in [8, 16]. Both are continuous. Let g(t) = u(t) - d(t). At t = 8: u = 0 and d = the full length \ell, so g(8) = -\ell < 0. At t = 16: g(16) = \ell - 0 > 0. By the IVT, g(t) = 0 for some t. (Picture both days happening at once: the two hikers must meet.)

11. 5.11Challenge Done

    (a) Prove that if f is continuous on an interval and never zero, then f has constant sign there. (b) Prove that a continuous function f : \R \to \R that takes only rational values is constant.

    Solution

    (a) If f(p) < 0 < f(q) for two points of the interval, the IVT gives a zero between them.

    (b) Suppose f(p) \ne f(q). Between the two values lies an irrational number k (Exercise 1.9). By the IVT, f takes the value k somewhere between p and q, contradicting that all its values are rational.

This is the idea calculus is named for. You know the derivative as a set of rules. This module defines it as a limit, computes it from that definition for the basic functions (the IB phrase is “differentiation from first principles”), and pins down exactly what differentiability means and what it implies.

6.1Two problems, one limit

The tangent problem. What is the slope of a curve y = f(x) at the point P = (a, f(a))? A line through P and a nearby point Q = (a + h, f(a+h)) on the curve, a secant, has slope

The tangent at P should be the limiting position of these secants as Q slides towards P, that is as h \to 0.

The velocity problem. If s(t) is the position of a moving object at time t, its average velocity over the time interval from t to t + h is \frac{s(t+h) - s(t)}{h}. Its velocity at the instant t should be the limit of these averages as h \to 0. You saw this for a falling stone in Module 4.

Both problems lead to the same limit, and Newton and Leibniz’s insight was that this one limit answers both.

6.2The definition

The derivative function f' assigns the number f'(x) to each x at which f is differentiable. Several notations are in use, and you need to read all of them:

Newton’s dot is standard in physics: in mechanics at NTNU you will write \dot x for velocity and \ddot x for acceleration.

The derivative has two readings (SL 5.1). Geometrically, f'(a) is the slope of the tangent line at (a, f(a)). Physically, it is the instantaneous rate of change of f with respect to x, measured in units of f per unit of x: metres per second, litres per minute, kroner per item.

6.3Differentiation from first principles

Constant functions have derivative 0, and f(x) = mx + c has \frac{f(x+h) - f(x)}{h} = m for every h \ne 0, so f'(x) = m. The first interesting case is x^2:

Everything the IB told you about differentiating trigonometric functions rests on these two limits, and those rest on radian measure.

From the limit laws, derivatives respect sums and constant multiples: (f + g)' = f' + g' and (cf)' = cf'. With the power rule this differentiates every polynomial (SL 5.3). Module 7 proves the product, quotient and chain rules.

6.4Differentiability and continuity

The converse is false. For f(x) = |x| at 0, the difference quotient \frac{|h|}{h} equals 1 for h > 0 and -1 for h < 0. Try the “|x|” option in the graph above: the secants from the right and from the left settle on different slopes.

Here is a gallery of the ways a continuous function can fail to be differentiable at 0. Each is worth sketching.

And a function that is differentiable but still surprising: g(x) = x^2\sin(1/x) with g(0) = 0. At 0 the quotient is \frac{h^2\sin(1/h)}{h} = h\sin(1/h) \to 0, so g'(0) = 0. In Module 7 you will see that g'(x) has no limit as x \to 0. So a derivative can exist everywhere without being a continuous function. The IB asks for an informal understanding of continuity and differentiability (AHL 5.12); this is what the formal version looks like.

6.5The best linear approximation

In words: near a, f is approximately the linear function h \mapsto f(a) + f'(a)h, with an error that is small even compared with h. If you zoom in on the graph of a differentiable function, it looks more and more like its tangent line. This gives the linear approximation

For example, with f(x) = \sqrt x and a = 4: \sqrt{4.1} \approx 2 + \frac{1}{4}(0.1) = 2.025, while the true value is 2.02485\ldots This point of view is the one that generalises: in several variables (your second semester) the derivative is the best linear approximation, and Taylor polynomials (Module 19) are the best polynomial approximations.

6.6Higher derivatives

If f' is itself differentiable, its derivative is the second derivative f'' = (f')', also written \frac{d^2y}{dx^2}. Continuing gives f''' and in general f^{(n)}, the nth derivative. If s(t) is position, then s' is velocity and s'' is acceleration (\ddot x in physics notation). andrederivert

Exercises

1. 6.1Fluency Done

   Differentiate from first principles: (a) 3x^2 - 5x + 2   (b) \dfrac{1}{x+2}   (c) \sqrt{2x+1}   (d) \dfrac1{x^2}.

   Solution

   (a) \frac{3(x+h)^2 - 5(x+h) + 2 - (3x^2 - 5x + 2)}{h} = \frac{6xh + 3h^2 - 5h}{h} = 6x + 3h - 5 \to 6x - 5.

   (b) \frac{1}{h}\left(\frac{1}{x+h+2} - \frac1{x+2}\right) = \frac{-1}{(x+h+2)(x+2)} \to -\frac{1}{(x+2)^2}.

   (c) \frac{\sqrt{2x+2h+1} - \sqrt{2x+1}}{h} = \frac{2h}{h\left(\sqrt{2x+2h+1}+\sqrt{2x+1}\right)} \to \frac{2}{2\sqrt{2x+1}} = \frac{1}{\sqrt{2x+1}}, for x > -\frac12.

   (d) \frac1h\left(\frac{1}{(x+h)^2} - \frac1{x^2}\right) = \frac{x^2 - (x+h)^2}{h\,x^2(x+h)^2} = \frac{-2x - h}{x^2(x+h)^2} \to \frac{-2x}{x^4} = -\frac{2}{x^3}.

2. 6.2Fluency Done

   Find the equations of the tangent and the normal to y = x^3 - 2x at the point where x = 1.

   Solution

   y(1) = -1 and y' = 3x^2 - 2, so y'(1) = 1. Tangent: y = -1 + (x - 1), i.e. y = x - 2. Normal: slope -1 through (1,-1), i.e. y = -x.

3. 6.3Fluency Done

   Each limit is a derivative f'(a) in disguise. Identify f and a and evaluate: (a) \displaystyle\lim_{h\to0}\frac{(2+h)^5 - 32}{h}   (b) \displaystyle\lim_{x\to\pi/3}\frac{\cos x - \frac12}{x - \frac\pi3}   (c) \displaystyle\lim_{h\to0}\frac{\sqrt{9+h} - 3}{h}.

   Solution

   (a) f(x) = x^5, a = 2: 5\cdot2^4 = 80. (b) f = \cos, a = \frac\pi3: -\sin\frac\pi3 = -\frac{\sqrt3}{2}. (c) f = \sqrt{\ }, a = 9: \frac{1}{2\sqrt9} = \frac16 (compare the direct computation in Module 4).

4. 6.4Fluency Done

   Where is f(x) = |x^2 - 4| not differentiable? Find the one-sided derivatives there.

   Solution

   f(x) = x^2 - 4 for |x| \ge 2 and 4 - x^2 for |x| < 2. Away from \pm2 it is a polynomial, hence differentiable. At x = 2: from the right the slope is 2x = 4, from the left -2x = -4. At x = -2: from the right -2x = 4, from the left 2x = -4. So f is not differentiable at x = \pm 2 (corners where the parabola is reflected).

5. 6.5Fluency Done

   A tank drains so that it holds V(t) = (20 - t)^2 litres after t minutes, 0 \le t \le 20. Find the average rate of change of V over the first 5 minutes and the instantaneous rate at t = 5, with units, and interpret the signs.

   Solution

   Average: \frac{V(5) - V(0)}{5} = \frac{225 - 400}{5} = -35 litres per minute. Instantaneous: V'(t) = -2(20 - t), so V'(5) = -30 litres per minute. Both are negative because the volume is decreasing. At t = 5 the tank is draining at 30 L/min, more slowly than its average over the first five minutes, because the outflow slows as the tank empties.

6. 6.6Understanding Done

   Prove by induction that \dfrac{d^n}{dx^n}\sin x = \sin\left(x + \dfrac{n\pi}{2}\right) for all n \in \N. (You may use the first-principles result that \frac{d}{dx}\sin(x + c) = \cos(x + c) for any constant c; the proof is identical to Theorem 6.5.)

   Solution

   For n = 1: (\sin x)' = \cos x = \sin(x + \frac\pi2). Suppose the formula holds for n. Then

   \frac{d^{n+1}}{dx^{n+1}}\sin x = \frac{d}{dx}\sin\left(x + \tfrac{n\pi}{2}\right) = \cos\left(x + \tfrac{n\pi}2\right) = \sin\left(x + \tfrac{n\pi}{2} + \tfrac\pi2\right) = \sin\left(x + \tfrac{(n+1)\pi}{2}\right).
7. 6.7Understanding Done

   Let f(x) = x|x|. Show that f is differentiable everywhere with f'(x) = 2|x|, but that f' is not differentiable at 0.

   Solution

   For x > 0, f(x) = x^2 and f'(x) = 2x; for x < 0, f(x) = -x^2 and f'(x) = -2x. At 0: \frac{f(h) - f(0)}{h} = \frac{h|h|}{h} = |h| \to 0. So f'(x) = 2|x| for all x, which has a corner at 0. A function can be differentiable once without being differentiable twice.

8. 6.8Understanding Done

   Show that f(x) = x^{1/3} is continuous at 0 but not differentiable there. Describe the tangent line at the origin.

   Solution

   |x^{1/3}| < \eps whenever |x| < \eps^3, so f is continuous at 0. The difference quotient is \frac{h^{1/3}}{h} = h^{-2/3} \to +\infty as h \to 0 from either side. The curve has a vertical tangent: the line x = 0.

9. 6.9Understanding Done

   (a) Suppose f is differentiable at a. Prove that \displaystyle\lim_{h\to0}\frac{f(a+h) - f(a-h)}{2h} = f'(a). (b) Show that this limit can exist when f is not differentiable at a. (c) For f = \sin and a = 1, compare the forward quotient \frac{f(a+h)-f(a)}{h} and the symmetric quotient with \cos 1 when h = 0.1.

   Solution

   (a) \frac{f(a+h) - f(a-h)}{2h} = \frac12\cdot\frac{f(a+h) - f(a)}{h} + \frac12\cdot\frac{f(a - h) - f(a)}{-h} \to \frac12 f'(a) + \frac12 f'(a).

   (b) f(x) = |x|, a = 0: the symmetric quotient is \frac{|h| - |-h|}{2h} = 0 for every h, so the limit is 0, but f'(0) does not exist.

   (c) Forward: \frac{\sin1.1 - \sin1}{0.1} \approx 0.49736. Symmetric: \frac{\sin 1.1 - \sin 0.9}{0.2} \approx 0.53940. True value \cos 1 \approx 0.54030. The errors are about 0.043 and 0.0009: the symmetric quotient is far more accurate. Module 19 explains why (its error is proportional to h^2, not h). Numerical software uses it for exactly this reason.

10. 6.10Challenge Done

    Suppose |f(x)| \le x^2 for all x. Prove that f is differentiable at 0 and f'(0) = 0.

    Solution

    First |f(0)| \le 0, so f(0) = 0. Then \left|\frac{f(h) - f(0)}{h}\right| = \frac{|f(h)|}{|h|} \le |h| \to 0, so the difference quotient tends to 0 by squeezing. This covers x^2\sin(1/x), x^2 D(x) with D the Dirichlet function, and many other strange functions at once.

11. 6.11Challenge Done

    Suppose f(x + y) = f(x)f(y) for all x, y, that f is not the zero function, and that f is differentiable at 0 with f'(0) = k. Prove that f(0) = 1 and that f is differentiable everywhere with f'(x) = k\,f(x).

    Solution

    Choose x_0 with f(x_0) \ne 0. Then f(x_0) = f(x_0 + 0) = f(x_0)f(0), so f(0) = 1. For any x,

    \frac{f(x+h) - f(x)}{h} = \frac{f(x)f(h) - f(x)}{h} = f(x)\cdot\frac{f(h) - f(0)}{h} \to f(x)\,f'(0) = k\,f(x).

    So a function that turns sums into products has a derivative proportional to itself. Module 9 shows that this is the defining property of exponential functions.

12. 6.12Challenge Done

    Let g(x) = x^2\sin(1/x) for x \ne 0 and g(0) = 0. Using only the definition, show that g is differentiable at 0. Then show that the function x\sin(1/x) (with value 0 at 0) is not differentiable at 0.

    Solution

    \frac{g(h) - g(0)}{h} = h\sin(1/h), and |h\sin(1/h)| \le |h| \to 0. So g'(0) = 0.

    For k(x) = x\sin(1/x): \frac{k(h) - k(0)}{h} = \sin(1/h), which has no limit as h \to 0 (Example 4.9). One extra factor of x makes the difference between differentiable and not.

With the definition in hand we can now prove the rules that let you differentiate almost anything without going back to limits: the product, quotient and chain rules. You used all three in IB. Here you will see why they are true, and why the “obvious” proof of the chain rule has a hole in it.

7.1Sums and constant multiples

Combined with the power rule this differentiates every polynomial: \frac{d}{dx}(4x^3 - 7x + 2) = 12x^2 - 7.

7.2The product rule

A picture explains the shape of the formula. Think of f and g as the sides of a rectangle of area fg. If the sides grow by \Delta f and \Delta g, the area grows by f\,\Delta g + g\,\Delta f + \Delta f\,\Delta g. Divided by \Delta x, the first two terms give the product rule, and the corner piece \Delta f\,\Delta g is of second order and disappears in the limit. Leibniz wrote this as d(uv) = u\,dv + v\,du.

7.3The quotient rule

This is the full statement of SL 5.3. Rational exponents, such as x^{3/2}, come in Module 8, and real exponents in Module 9.

7.4The chain rule

The tempting proof is to write

and let h \to 0. The flaw is that g(a+h) - g(a) may be zero for values of h arbitrarily close to 0, and then the first fraction is meaningless. This really happens: for g(x) = x^2\sin(1/x) (with g(0) = 0) we have g(h) = 0 at every h = \frac{1}{n\pi}. The repair is to avoid dividing by g(a+h) - g(a) at all.

In practice, differentiate from the outside in, multiplying by the derivative of each inner layer:

7.5All six trigonometric functions

A memory aid: the “co-” functions (\cos, \csc, \cot) are the ones with a minus sign. These are the AHL 5.15 derivatives of \tan x, \sec x, \csc x and \cot x.

7.6Using the rules together

Exercises

1. 7.1Fluency Done

   Differentiate: (a) (3x^2 - 1)^4   (b) x^3\cos 2x   (c) \dfrac{\sin x}{1 + \cos x}   (d) \sec(x^2)   (e) \cot^3(5x)   (f) \dfrac{x}{(x^2+1)^2}

   Solution

   (a) 4(3x^2-1)^3\cdot6x = 24x(3x^2-1)^3.

   (b) 3x^2\cos2x - 2x^3\sin 2x.

   (c) \frac{\cos x(1+\cos x) + \sin^2 x}{(1+\cos x)^2} = \frac{\cos x + 1}{(1+\cos x)^2} = \frac{1}{1 + \cos x}.

   (d) 2x\sec(x^2)\tan(x^2).

   (e) 3\cot^2(5x)\cdot\left(-\csc^2(5x)\right)\cdot5 = -15\cot^2(5x)\csc^2(5x).

   (f) \frac{(x^2+1)^2 - x\cdot2(x^2+1)\cdot2x}{(x^2+1)^4} = \frac{(x^2+1) - 4x^2}{(x^2+1)^3} = \frac{1 - 3x^2}{(x^2+1)^3}.

2. 7.2Fluency Done

   Find the equation of the tangent to y = \tan x at x = \frac\pi4.

   Solution

   y = 1 and y' = \sec^2\frac\pi4 = 2. Tangent: y = 1 + 2\left(x - \frac\pi4\right) = 2x + 1 - \frac\pi2.

3. 7.3Fluency Done

   The curve y = \dfrac{ax + b}{x - 1} passes through (2, 5) and has gradient -3 there. Find a and b.

   Solution

   Through (2,5): 2a + b = 5. The derivative is \frac{a(x-1) - (ax+b)}{(x-1)^2} = \frac{-a-b}{(x-1)^2}, so at x = 2: -a - b = -3. Solving: a = 2, b = 1.

4. 7.4Fluency Done

   Given f(2) = 3, f'(2) = -1, g(2) = 5, g'(2) = 4 and g'(3) = 7, find (fg)'(2), \left(\frac fg\right)'(2), (g\circ f)'(2) and (f^2)'(2).

   Solution

   (fg)'(2) = (-1)(5) + 3(4) = 7. \ \left(\frac fg\right)'(2) = \frac{(-1)(5) - 3(4)}{25} = -\frac{17}{25}. \ (g\circ f)'(2) = g'(f(2))f'(2) = g'(3)(-1) = -7. \ (f^2)'(2) = 2f(2)f'(2) = -6.

5. 7.5Fluency Done

   Show that y = \sin 2x + \cos 2x satisfies y'' + 4y = 0.

   Solution

   y' = 2\cos 2x - 2\sin 2x and y'' = -4\sin 2x - 4\cos 2x = -4y. This is the equation of simple harmonic motion, which you will solve in general in your first year at NTNU.

6. 7.6Understanding Done

   Derive the quotient rule directly from the definition of the derivative, without using the product rule.

   Solution

   Put everything over a common denominator and add and subtract f(x)g(x):

   \frac1h\left(\frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}\right) = \frac{\big(f(x+h) - f(x)\big)g(x) - f(x)\big(g(x+h) - g(x)\big)}{h\,g(x+h)g(x)} \to \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2},

   using g(x+h) \to g(x) \ne 0.

7. 7.7Understanding Done

   Prove \frac{d}{dx}x^n = nx^{n-1} for n \in \N by induction, using only the product rule and \frac{d}{dx}x = 1.

   Solution

   True for n = 1. If (x^n)' = nx^{n-1}, then (x^{n+1})' = (x\cdot x^n)' = 1\cdot x^n + x\cdot nx^{n-1} = (n+1)x^n.

8. 7.8Understanding Done

   Prove by induction that \dfrac{d^n}{dx^n}\left(\dfrac1x\right) = \dfrac{(-1)^n\,n!}{x^{n+1}} for all n \in \N.

   Solution

   For n = 1: -\frac{1}{x^2}, correct. If the formula holds for n, then by the power rule for integers

   \frac{d}{dx}\Big[(-1)^n n!\,x^{-(n+1)}\Big] = (-1)^n n!\cdot\big(-(n+1)\big)x^{-(n+2)} = \frac{(-1)^{n+1}(n+1)!}{x^{n+2}}.
9. 7.9Understanding Done

   Let g(x) = x^2\sin(1/x) for x \ne 0 and g(0) = 0. Find g'(x) for x \ne 0, and show that \lim_{x\to0}g'(x) does not exist even though g'(0) = 0.

   Solution

   For x \ne 0, by the product and chain rules, g'(x) = 2x\sin(1/x) + x^2\cos(1/x)\cdot\left(-\frac1{x^2}\right) = 2x\sin(1/x) - \cos(1/x). The first term tends to 0, but \cos(1/x) has no limit as x \to 0 (same argument as for \sin(1/x)). So g'(x) has no limit, while g'(0) = 0 from Module 6. g is differentiable everywhere, but g' is not continuous at 0.

10. 7.10Understanding Done

    For the function g of the previous exercise, show that g\left(\frac{1}{n\pi}\right) = 0 for every n \in \N. Explain why this breaks the “tempting proof” of the chain rule when the inner function is g and a = 0.

    Solution

    g\left(\frac1{n\pi}\right) = \frac{1}{n^2\pi^2}\sin(n\pi) = 0 = g(0). So g(0 + h) - g(0) = 0 for h = \frac1{n\pi}, values of h arbitrarily close to 0. The tempting proof divides by g(a+h) - g(a), which is 0 at all these h, so its first fraction is undefined infinitely often near h = 0 and its limit cannot even be formed.

11. 7.11Challenge Done

    Prove Leibniz’s rule (fg)^{(n)} = \sum_{k=0}^n\binom nk f^{(k)}g^{(n-k)} by induction, and use it to find the 10th derivative of x^2\sin x.

    Solution

    For n = 1 it is the product rule. Assume it for n and differentiate each term with the product rule:

    (fg)^{(n+1)} = \sum_{k=0}^n\binom nk\left(f^{(k+1)}g^{(n-k)} + f^{(k)}g^{(n+1-k)}\right) = \sum_{k=0}^{n+1}\left[\binom n{k-1} + \binom nk\right]f^{(k)}g^{(n+1-k)},

    and \binom n{k-1} + \binom nk = \binom{n+1}k (Pascal’s rule).

    With f = x^2 only k = 0, 1, 2 contribute:

    \frac{d^{10}}{dx^{10}}(x^2\sin x) = x^2\sin^{(10)}x + 10\cdot2x\sin^{(9)}x + 45\cdot2\sin^{(8)}x = -x^2\sin x + 20x\cos x + 90\sin x,

    using \sin^{(n)}x = \sin(x + \frac{n\pi}2).

12. 7.12Challenge Done

    Prove that the derivative of a differentiable even function is odd, and the derivative of a differentiable odd function is even. Prove also that the derivative of a differentiable periodic function is periodic with the same period.

    Solution

    If f(-x) = f(x) for all x, differentiate both sides with the chain rule: -f'(-x) = f'(x), so f' is odd. If f(-x) = -f(x), then -f'(-x) = -f'(x), so f' is even. If f(x + p) = f(x), then f'(x+p) = f'(x). (Check: \cos is even and \cos' = -\sin is odd.)

If y = f(x) and we know \frac{dy}{dx}, what is \frac{dx}{dy}? The answer, its reciprocal, is the key to differentiating every inverse function: roots, the inverse trigonometric functions, and in Module 9 the logarithm. The same idea, differentiating an equation without solving it first, gives implicit differentiation.

8.1The derivative of an inverse function

The graph of f^{-1} is the graph of f reflected in the line y = x (Module 2). Reflection swaps horizontal and vertical, so it swaps “rise” and “run”: a tangent of slope m at (a, b) on the graph of f becomes a tangent of slope \frac1m at (b, a) on the graph of f^{-1}.

The hypothesis f'(a) \ne 0 matters. The function x^3 has a horizontal tangent at 0, so its inverse \sqrt[3]{x} has a vertical tangent at 0 and is not differentiable there (compare Module 6).

8.2Rational powers

This completes SL 5.6’s “derivative of x^n, n \in \Q”. Where the power is defined for negative x (for example x^{1/3} or x^{2/3}) the same formula holds for x < 0 too, by symmetry. Module 9 extends it to irrational exponents.

8.3The inverse trigonometric functions

Two things are worth noticing. The derivatives of \arcsin and \arccos differ only in sign, so \arcsin x + \arccos x has derivative 0; Module 10 shows it is the constant \frac\pi2. And these transcendental functions have algebraic derivatives, which is why they turn up so often as answers to integrals.

Read backwards, these give \int\frac{dx}{\sqrt{a^2-x^2}} = \arcsin\frac xa + C and \int\frac{dx}{a^2+x^2} = \frac1a\arctan\frac xa + C in Module 15.

8.4Implicit differentiation

Many curves are given by an equation in x and y rather than a formula y = f(x): circles, ellipses, and curves like x^3 + y^3 = 6xy. Near most points such a curve is still the graph of a differentiable function y(x), even if we cannot write it down. The method (AHL 5.14) is:

1. differentiate both sides of the equation with respect to x, treating y as a function of x and using the chain rule wherever y appears;
2. solve the resulting equation for y'.

The chain rule gives, for example, \frac{d}{dx}y^3 = 3y^2y', \ \frac{d}{dx}\sin y = \cos y\cdot y', and by the product rule \frac{d}{dx}(xy) = y + xy'.

Exercises

1. 8.1Fluency Done

   Differentiate: (a) \arcsin(3x)   (b) \arctan(x^2)   (c) \arccos\frac1x for x > 1   (d) (\arcsin x)^2   (e) \sqrt[3]{x^2+1}   (f) \sqrt x\,\arctan\sqrt x

   Solution

   (a) \frac{3}{\sqrt{1-9x^2}}   (b) \frac{2x}{1+x^4}

   (c) -\frac{1}{\sqrt{1 - 1/x^2}}\cdot\left(-\frac1{x^2}\right) = \frac{1}{x^2\sqrt{(x^2-1)/x^2}} = \frac{1}{x\sqrt{x^2-1}} (using x > 0).

   (d) \frac{2\arcsin x}{\sqrt{1-x^2}}   (e) \frac{2x}{3}(x^2+1)^{-2/3}

   (f) \frac{\arctan\sqrt x}{2\sqrt x} + \sqrt x\cdot\frac{1}{1 + x}\cdot\frac{1}{2\sqrt x} = \frac{\arctan\sqrt x}{2\sqrt x} + \frac{1}{2(1+x)}.

2. 8.2Fluency Done

   Find \frac{dy}{dx}: (a) x^2 + xy + y^2 = 7   (b) \sin(xy) = x   (c) \sqrt x + \sqrt y = 3.

   Solution

   (a) 2x + y + xy' + 2yy' = 0, so y' = -\frac{2x + y}{x + 2y}.

   (b) \cos(xy)(y + xy') = 1, so y' = \frac{\sec(xy) - y}{x}.

   (c) \frac{1}{2\sqrt x} + \frac{y'}{2\sqrt y} = 0, so y' = -\frac{\sqrt y}{\sqrt x}.

3. 8.3Fluency Done

   For the curve x^2 + xy + y^2 = 7: find the tangent at (1, 2), and find the points where the tangent is horizontal.

   Solution

   At (1,2): y' = -\frac{2 + 2}{1 + 4} = -\frac45, so the tangent is y - 2 = -\frac45(x - 1).

   Horizontal tangent: y' = 0 needs 2x + y = 0, i.e. y = -2x. Substituting: x^2 - 2x^2 + 4x^2 = 3x^2 = 7, so x = \pm\sqrt{7/3}, giving the points \left(\sqrt{7/3}, -2\sqrt{7/3}\right) and \left(-\sqrt{7/3}, 2\sqrt{7/3}\right). (At these points x + 2y \ne 0, so the formula for y' is valid.)

4. 8.4Fluency Done

   (a) For f(x) = x^3 + x + 1, find (f^{-1})'(3) and (f^{-1})'(1). (b) For g(x) = 2x + \cos x, explain why g has an inverse and find (g^{-1})'(1).

   Solution

   (a) f(1) = 3 and f'(1) = 4, so (f^{-1})'(3) = \frac14. \ f(0) = 1 and f'(0) = 1, so (f^{-1})'(1) = 1.

   (b) g'(x) = 2 - \sin x \ge 1 > 0, so g is strictly increasing (Module 10 proves that a positive derivative implies this) and has an inverse. g(0) = 1 and g'(0) = 2, so (g^{-1})'(1) = \frac12.

5. 8.5Understanding Done

   Show that the derivative of \arcsin x + \arccos x is 0 on (-1, 1). What do you expect the function to be?

   Solution

   \frac{1}{\sqrt{1-x^2}} - \frac1{\sqrt{1-x^2}} = 0. We expect a constant, and its value at 0 is 0 + \frac\pi2. So \arcsin x + \arccos x = \frac\pi2; Module 10 proves that zero derivative on an interval forces a constant. (Geometrically: in a right-angled triangle with hypotenuse 1 and a side x, the two acute angles are \arcsin x and \arccos x.)

6. 8.6Understanding Done

   For the curve x^3 + y^3 = 1, find y' and y'' in terms of x and y, simplifying with the equation of the curve.

   Solution

   3x^2 + 3y^2y' = 0, so y' = -\frac{x^2}{y^2}. Then

   y'' = -\frac{2xy^2 - x^2\cdot 2yy'}{y^4} = -\frac{2xy^2 + 2x^4/y}{y^4} = -\frac{2x(y^3 + x^3)}{y^5} = -\frac{2x}{y^5}.
7. 8.7Understanding Done

   Show that h(x) = \arctan x + \arctan\frac1x has h'(x) = 0 for every x \ne 0. Evaluate h(1) and h(-1). What does this tell you about functions with zero derivative?

   Solution

   h'(x) = \frac{1}{1+x^2} + \frac{1}{1 + 1/x^2}\cdot\left(-\frac{1}{x^2}\right) = \frac{1}{1 + x^2} - \frac{1}{x^2 + 1} = 0.

   h(1) = \frac\pi4 + \frac\pi4 = \frac\pi2 and h(-1) = -\frac\pi2. So h is not constant, although its derivative is zero wherever it is defined. The domain \R\setminus\{0\} is not an interval, and “zero derivative implies constant” only holds on intervals (Module 10). On each half-line, h is constant.

8. 8.8Understanding Done

   Let f(x) = \tan x on (-\frac\pi2, \frac\pi2). Using only the inverse function rule, find the derivative of \arctan at 1 and at \sqrt3, and check your answers against the formula \frac{1}{1+x^2}.

   Solution

   \tan\frac\pi4 = 1 and \tan'\frac\pi4 = \sec^2\frac\pi4 = 2, so (\arctan)'(1) = \frac12 = \frac{1}{1+1}. \ \tan\frac\pi3 = \sqrt3 and \sec^2\frac\pi3 = 4, so (\arctan)'(\sqrt3) = \frac14 = \frac{1}{1+3}.

9. 8.9Challenge Done

   Let f(x) = x^5 + x and let g = f^{-1}. Explain why g : \R \to \R exists and is differentiable, and find g'(2) and g''(2).

   Solution

   f'(x) = 5x^4 + 1 > 0, so f is strictly increasing; it is continuous and unbounded in both directions, so by the IVT its range is \R. The inverse function rule applies everywhere since f' \ne 0.

   f(1) = 2, so g'(2) = \frac{1}{f'(1)} = \frac16. Differentiating g'(y) = \frac{1}{f'(g(y))} with the chain rule:

   g''(y) = -\frac{f''(g(y))\,g'(y)}{f'(g(y))^2} = -\frac{f''(x)}{f'(x)^3}\quad\text{where } x = g(y).

   With x = 1: f''(1) = 20 and f'(1) = 6, so g''(2) = -\frac{20}{216} = -\frac{5}{54}.

10. 8.10Challenge Done

    Find the points where the folium of Descartes x^3 + y^3 = 6xy has a horizontal tangent.

    Solution

    From Example 8.8, y' = 0 needs 2y - x^2 = 0 with y^2 - 2x \ne 0. Substitute y = \frac{x^2}2: x^3 + \frac{x^6}8 = 3x^3, so x^3\left(\frac{x^3}{8} - 2\right) = 0. The root x = 0 gives the origin, where y^2 - 2x = 0 too and the formula fails (the curve crosses itself). The other root is x^3 = 16, i.e. x = 2^{4/3}, with y = \frac{x^2}{2} = 2^{5/3}. There y^2 - 2x = 2^{10/3} - 2^{7/3} \ne 0. So the only horizontal tangent is at \left(2^{4/3}, 2^{5/3}\right) \approx (2.52, 3.17).

11. 8.11Challenge Done

    Show that for the astroid x^{2/3} + y^{2/3} = 1, the part of any tangent line (at a point with x, y > 0) that lies between the coordinate axes has length 1.

    Solution

    Differentiating, \frac23x^{-1/3} + \frac23y^{-1/3}y' = 0, so y' = -\left(\frac yx\right)^{1/3}. At (x_0, y_0) the tangent meets the x-axis where

    x = x_0 + y_0\left(\tfrac{x_0}{y_0}\right)^{1/3} = x_0^{1/3}\left(x_0^{2/3} + y_0^{2/3}\right) = x_0^{1/3},

    and similarly the y-axis at y = y_0^{1/3}. The segment between these intercepts has length \sqrt{x_0^{2/3} + y_0^{2/3}} = 1. (A ladder of length 1 sliding down a wall sweeps out exactly this curve; see Module 13.)

In IB you were told that e^x is its own derivative and that \ln x has derivative \frac1x. Here we earn those facts. We define a^x for every real x, show that the derivative of any exponential function is proportional to the function itself, find the one base for which the constant of proportionality is 1, and check that it is the same e that appeared in Module 3.

9.1Powers with real exponents

Fix a base a > 0. Integer powers need no comment. Module 5 gave us nth roots, so a^{1/n} is defined, and then a^{p/q} = \left(a^{1/q}\right)^p. The laws of exponents

hold for all rational r and s; they follow from the integer case by straightforward algebra. What does 2^{\sqrt2} or 3^\pi mean? For a > 1 the function q \mapsto a^q is increasing on the rationals, so it is natural to define, for real x,

The set is bounded above (by a^m for any integer m \ge x), so completeness provides the supremum. For 0 < a < 1 put a^x = 1/(1/a)^x, and 1^x = 1.

9.2The derivative of an exponential function

From first principles, the laws of exponents give

The second factor does not involve x at all. So everything depends on one number per base:

the slope of y = a^x at x = 0. If this limit exists, then for every x

The derivative of an exponential function is a constant multiple of the function itself. Numerically:

The table suggests L(2) \approx 0.693 and L(3) \approx 1.099, and that the base with L(a) = 1 lies between 2.7 and 2.8. But does the limit exist at all?

9.3The number e

Since L(2^t) = t\,L(2) with L(2) > 0, as t runs through \R the slope L(2^t) runs through every real number, each exactly once. Exactly one value, t = \frac{1}{L(2)}, gives slope 1. Call that base e:

So e is the unique base whose exponential has slope exactly 1 where it crosses the y-axis. The table puts it between 2.7 and 2.8. In Section 9.7 we show it is the same number \lim(1 + \frac1n)^n as in Module 3. With the chain rule:

In Module 10 we will prove the converse: e^x is the only function with f' = f and f(0) = 1. That makes e^x the basic solution of the simplest differential equation, which is why it appears in every model of growth, decay, oscillation and diffusion you will meet in physics.

9.4The natural logarithm

e^x is continuous and strictly increasing from \R onto (0,\infty), so it has an inverse function, the natural logarithm \ln : (0,\infty) \to \R:

den naturlige logaritmen The laws \ln(xy) = \ln x + \ln y, \ \ln(x^r) = r\ln x, \ \ln 1 = 0 and \ln e = 1 follow from the laws of exponents.

By the chain rule, \frac{d}{dx}\ln|g(x)| = \frac{g'(x)}{g(x)}. A transcendental function with the simplest possible algebraic derivative: this fills the one gap in the power rule. Every power x^n with n \ne -1 is the derivative of \frac{x^{n+1}}{n+1}, and \ln|x| is exactly the missing function whose derivative is x^{-1}.

9.5Other bases and real powers

Since e^{x\ln a} = \left(e^{\ln a}\right)^x = a^x, every exponential is an exponential of base e in disguise:

So the mysterious constant is L(a) = \ln a: \ L(2) = \ln 2 = 0.693147\ldots and L(3) = \ln 3 = 1.098612\ldots, matching the table. For logarithms to base a, \ \log_a x = \frac{\ln x}{\ln a}, so

These are the AHL 5.15 derivatives of a^x and \log_a x. Finally, for x > 0 and any real r, x^r = e^{r\ln x}, so

The power rule now holds for every real exponent, x^{\pi} and x^{\sqrt2} included.

9.6Logarithmic differentiation

To differentiate a complicated product, quotient or power, take logarithms first, then differentiate implicitly.

9.7The two definitions of e agree

More generally \left(1 + \frac xn\right)^n \to e^x for every real x (Exercise 9.10). With interest rate x compounded n times a year, this is the balance after one year, and e^x is the result of compounding continuously.

9.8Beyond IB: hyperbolic functions

Exercises

1. 9.1Fluency Done

   Differentiate: (a) e^{3x^2}   (b) x^2\ln x   (c) \ln(\sec x)   (d) 5^x   (e) \log_2(x^2 + 1)   (f) e^{-x}\sin 2x   (g) \ln\dfrac{1+x}{1-x}

   Solution

   (a) 6xe^{3x^2}   (b) 2x\ln x + x   (c) \frac{\sec x\tan x}{\sec x} = \tan x   (d) 5^x\ln5   (e) \frac{2x}{(x^2+1)\ln2}   (f) e^{-x}(2\cos2x - \sin2x)

   (g) \ln(1+x) - \ln(1-x) has derivative \frac1{1+x} + \frac1{1-x} = \frac{2}{1-x^2}.

2. 9.2Fluency Done

   Use logarithmic differentiation: (a) y = x^{\sin x}   (b) y = \dfrac{x^3(x+1)^4}{\sqrt{x^2+1}}   (c) y = (\ln x)^x for x > 1.

   Solution

   (a) \ln y = \sin x\ln x, so y' = x^{\sin x}\left(\cos x\ln x + \frac{\sin x}x\right).

   (b) y' = y\left(\frac3x + \frac{4}{x+1} - \frac{x}{x^2+1}\right).

   (c) \ln y = x\ln(\ln x), so \frac{y'}y = \ln(\ln x) + x\cdot\frac{1}{\ln x}\cdot\frac1x and y' = (\ln x)^x\left(\ln(\ln x) + \frac{1}{\ln x}\right).

3. 9.3Fluency Done

   Find the equation of the tangent to y = \ln x that passes through the origin.

   Solution

   The tangent at (a, \ln a) is y - \ln a = \frac1a(x - a). It passes through (0,0) when -\ln a = -1, so a = e. The tangent is y = \frac xe, touching at (e, 1).

4. 9.4Fluency Done

   A population grows as P(t) = 2000e^{0.03t} (t in years). Find the growth rate when t = 10, and when P = 5000.

   Solution

   P'(t) = 0.03\cdot2000e^{0.03t} = 0.03P. At t = 10: 60e^{0.3} \approx 81 per year. When P = 5000: 0.03 \cdot 5000 = 150 per year. The rate is proportional to the current size, which is what “exponential growth” means.

5. 9.5Understanding Done

   Working directly from L(a) = \lim_{h\to0}\frac{a^h-1}h, prove that L(1/a) = -L(a).

   Solution

   \frac{(1/a)^h - 1}{h} = \frac{a^{-h} - 1}{h} = -\frac{a^{-h} - 1}{-h}. As h \to 0, also -h \to 0, so this tends to -L(a). (Equivalently, L(a) + L(1/a) = L(1) = 0 by Proposition 9.2.)

6. 9.6Understanding Done

   Show that y = e^{-x^2/2} satisfies y' = -xy, and find the points where y'' = 0.

   Solution

   y' = -xe^{-x^2/2} = -xy. Then y'' = -y - xy' = -y + x^2y = (x^2 - 1)e^{-x^2/2}, which vanishes at x = \pm1. These are the inflexion points of the bell curve; for the normal distribution they sit one standard deviation from the mean.

7. 9.7Understanding Done

   Prove that (\cosh x)' = \sinh x, (\sinh x)' = \cosh x and \cosh^2x - \sinh^2 x = 1. Then show that \sinh has inverse \operatorname{arsinh}x = \ln\left(x + \sqrt{x^2+1}\right), and find its derivative.

   Solution

   The derivatives come straight from (e^{\pm x})' = \pm e^{\pm x}. And \cosh^2 x - \sinh^2 x = \frac{(e^x + e^{-x})^2 - (e^x - e^{-x})^2}{4} = \frac{4}{4} = 1.

   Solve x = \frac{e^y - e^{-y}}2 for y: multiplying by 2e^y gives e^{2y} - 2xe^y - 1 = 0, a quadratic in e^y, so e^y = x \pm\sqrt{x^2+1}. Since e^y > 0 we need the plus sign, and y = \ln\left(x + \sqrt{x^2+1}\right). Its derivative is

   \frac{1 + \frac{x}{\sqrt{x^2+1}}}{x + \sqrt{x^2+1}} = \frac{1}{\sqrt{x^2+1}}.

   Compare (\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}.

8. 9.8Understanding Done

   Prove that \displaystyle\lim_{x\to0}\frac{\ln(1+x)}{x} = 1, and deduce that \displaystyle\lim_{x\to0}(1+x)^{1/x} = e.

   Solution

   \frac{\ln(1 + x) - \ln 1}{x} is a difference quotient of \ln at 1, so its limit is \ln'(1) = 1. Then (1+x)^{1/x} = e^{\ln(1+x)/x} \to e^1 by continuity of the exponential.

9. 9.9Understanding Done

   Find where f(x) = x^x (x > 0) has a horizontal tangent, and the value of f there.

   Solution

   f'(x) = x^x(\ln x + 1) = 0 exactly when \ln x = -1, i.e. x = \frac1e. There f = e^{-1/e} \approx 0.692. (Module 12 shows this is the global minimum.)

10. 9.10Challenge Done

    Prove that \left(1 + \frac xn\right)^n \to e^x as n \to \infty, for every real x.

    Solution

    For x = 0 it is trivial. Otherwise, for n > |x|, \ \ln\left(1 + \frac xn\right)^n = x\cdot\frac{\ln\left(1 + \frac xn\right) - \ln1}{x/n}. The fraction is a difference quotient of \ln at 1 with step \frac xn \to 0, so it tends to 1, and the whole expression tends to x. By continuity of the exponential, \left(1 + \frac xn\right)^n \to e^x.

11. 9.11Challenge Done

    Fill in Step 1 of Theorem 9.1: prove that for a > 0 and rationals p, q, \ a^{(p+q)/2} \le \frac{a^p + a^q}{2}, with equality only if p = q or a = 1.

    Solution

    Let u = a^{p/2} and v = a^{q/2}, both positive. Then \frac{a^p + a^q}{2} - a^{(p+q)/2} = \frac{u^2 + v^2}{2} - uv = \frac{(u - v)^2}{2} \ge 0, with equality only if u = v, which means a^{p/2} = a^{q/2}, i.e. p = q or a = 1.

12. 9.12Challenge Done

    Differentiate y = x^{x^x} for x > 0. (Here x^{x^x} means x^{(x^x)}.)

    Solution

    \ln y = x^x\ln x. Using (x^x)' = x^x(\ln x + 1) and the product rule,

    \frac{y'}{y} = x^x(\ln x + 1)\ln x + \frac{x^x}x, \qquad y' = x^{x^x}\,x^x\left(\ln^2 x + \ln x + \frac1x\right).

The derivative is defined one point at a time, yet we constantly use it to draw conclusions about whole intervals: a positive derivative means increasing, a zero derivative means constant. Those conclusions are not obvious, and they are false on sets with gaps. The bridge from points to intervals is the Mean Value Theorem. After the Fundamental Theorem, which it also proves, it is the most important theorem in this course.

10.1Local extrema and Fermat’s theorem

Fermat’s theorem says: interior local extrema can only occur at critical points. It does not say that every critical point is an extremum.

10.2Rolle’s theorem