---
title: "Limits and Continuity"
book: "High School Mathematics"
subject: math
language: en
chapter: 21
exercises: 9
source: https://one-course.com/books/math/2/en/chapter/21-limits-and-continuity
---

# Chapter 21 — Limits and Continuity

This chapter extends the notion of limit from [sequences](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) to [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) of a real variable, and introduces [continuity](#def-g12-limcont-continuity). The central result is the intermediate value theorem, which turns qualitative information (“$f$ is [continuous](#def-g12-limcont-continuity) and changes sign”) into the existence of solutions of [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation).

## 21.1 Limits of functions

Throughout, $f$ is a [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) defined on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) or a [union](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) of [intervals](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $D \subseteq \R$.

**Definition 21.1 (Limit at infinity).**

Let $f$ be defined on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $\intco{a}{+\infty}$ and $\ell \in \R$. We say that $f$ *tends to $\ell$ at $+\infty$*, written $\lim\limits_{x\to+\infty} f(x) = \ell$, if every open [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) containing $\ell$ contains all values $f(x)$ for $x$ large enough: for every $\varepsilon > 0$ there exists $A$ such that for all $x \geq A$, $\abs{f(x) - \ell} \leq \varepsilon$.

We say $f$ tends to $+\infty$ at $+\infty$ if for every $M \in \R$ there exists $A$ such that $f(x) \geq M$ for all $x \geq A$. Limits at $-\infty$ and limits equal to $-\infty$ are defined analogously.

**Definition 21.2 (Limit at a point).**

Let $a \in \R$ and let $f$ be defined near $a$ (except possibly at $a$). We say that $f$ tends to $\ell \in \R$ at $a$, written $\lim\limits_{x \to a} f(x) = \ell$, if for every $\varepsilon > 0$ there exists $\delta > 0$ such that for all $x \in D$ with $\abs{x - a} \leq \delta$, $\abs{f(x) - \ell} \leq \varepsilon$.

We say $f$ tends to $+\infty$ at $a$ if for every $M$ there is $\delta > 0$ such that $f(x) \geq M$ whenever $x \in D$ and $\abs{x - a} \leq \delta$. One-sided limits ($x \to a^+$, $x \to a^-$) restrict $x$ to $x > a$ or $x < a$.

**Example 21.3.**

$\lim\limits_{x\to+\infty} \frac{1}{x} = 0$, $\lim\limits_{x\to 0^+} \frac{1}{x} = +\infty$, $\lim\limits_{x\to 0^-} \frac{1}{x} = -\infty$. The [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $x \mapsto \frac1x$ has no (two-sided) limit at $0$.

**Definition 21.4 (Asymptotes).**

The line $y = \ell$ is a *horizontal asymptote* of the curve of $f$ at $+\infty$ (resp. $-\infty$) if $\lim_{x\to+\infty} f(x) = \ell$ (resp. at $-\infty$). The line $x = a$ is a *vertical asymptote* if $f$ has an infinite one-sided limit at $a$.

![The curve of f(x) = 2 + 1/x-1 with its horizontal asymptote y = 2 (at ±∈fty) and vertical asymptote x = 1.](https://one-course.com/images/onecourse/chapters/math-2/g12-limcont/fig-80a817492b00.svg)

*The curve of $f(x) = 2 + \frac{1}{x-1}$ with its [horizontal asymptote](#def-g12-limcont-asymptote) $y = 2$ (at $\pm\infty$) and [vertical asymptote](#def-g12-limcont-asymptote) $x = 1$.*

**Proposition 21.5 (Operations on limits).**

The rules of [Proposition 20.14](https://one-course.com/books/math/2/en/chapter/20-sequences#prop-g12-seq-operations) (sums, products, quotients) hold verbatim for limits of [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function), at a point or at infinity, with the same four indeterminate forms $(+\infty)+(-\infty)$, $0\times\infty$, $\frac{\infty}{\infty}$, $\frac{0}{0}$.

**Proof.** The proofs are identical to the [sequence](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) case, replacing “for $n \geq N$” by “for $x \geq A$” or “for $\abs{x-a} \leq \delta$”. ∎

**Theorem 21.6 (Limit of a composition).**

Let $f, g$ be [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) and $a, b, c$ be [real numbers](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) or $\pm\infty$. If $\lim\limits_{x \to a} f(x) = b$ and $\lim\limits_{y \to b} g(y) = c$, then

$$
\lim_{x \to a} g\bigl(f(x)\bigr) = c .
$$

The same statement holds with a [sequence](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) in place of $f$: if $u_n \to b$ and $\lim_{y\to b} g(y) = c$, then $g(u_n) \to c$.

**Proof.** Take $b, c$ finite (the other cases are analogous). Let $\varepsilon > 0$. There is $\delta > 0$ such that $\abs{y - b} \leq \delta$ implies $\abs{g(y) - c} \leq \varepsilon$; and there is $\delta' > 0$ such that $\abs{x - a} \leq \delta'$ implies $\abs{f(x) - b} \leq \delta$. Chaining the two gives $\abs{g(f(x)) - c} \leq \varepsilon$ for $\abs{x - a} \leq \delta'$. ∎

**Theorem 21.7 (Comparison and squeeze).**

Let $f, g, h$ be [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) and $a \in \R \cup \{\pm\infty\}$.

1. If $f \leq g$ near $a$ and $f(x) \to +\infty$ as $x \to a$ , then $g(x) \to +\infty$ .
2. If $f \leq g \leq h$ near $a$ and $f, h$ tend to the same finite limit $\ell$ at $a$ , then $g(x) \to \ell$ as $x \to a$ .

**Proof.** Same argument as for [sequences](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) ([Theorem 20.16](https://one-course.com/books/math/2/en/chapter/20-sequences#thm-g12-seq-squeeze)). ∎

**Method 21.8 (Computing limits in practice).**

- Polynomials and rational [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) at $\pm\infty$ : factor out the highest power of $x$ ; the limit is that of the ratio of leading terms.
- $\frac{0}{0}$ at a point $a$ : factor $(x-a)$ out of numerator and denominator, or recognize a [derivative](https://one-course.com/books/math/2/en/chapter/12-differentiation-a-first-course#def-g11-deriv-derivative) $\lim_{x\to a}\frac{f(x)-f(a)}{x-a} = f'(a)$ ( [Chapter 22](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#ch-g12-deriv) ).
- Square roots: multiply by the conjugate expression.
- Oscillating factors ( $\cos$ , $\sin$ ): use the squeeze theorem.

## 21.2 Continuity

**Definition 21.9 (Continuity).**

Let $f$ be defined on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $I$ and $a \in I$. The [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $f$ is *continuous at $a$* if $\lim\limits_{x \to a} f(x) = f(a)$. It is *continuous on $I$* if it is continuous at every point of $I$.

Intuitively, the [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) of a [continuous](#def-g12-limcont-continuity) [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) can be drawn without lifting the pen.

**Proposition 21.10 (Continuity of usual functions).**

Polynomials, rational [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function), $x \mapsto \sqrt{x}$, $x \mapsto \abs{x}$, $\cos$, $\sin$, $\exp$ and $\ln$ are [continuous](#def-g12-limcont-continuity) on every [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) contained in their [domain](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function). Sums, products, quotients (where defined) and compositions of [continuous](#def-g12-limcont-continuity) [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) are [continuous](#def-g12-limcont-continuity).

**Partial proof.** Stability under sums, products, quotients and composition follows from the corresponding theorems on limits. [Continuity](#def-g12-limcont-continuity) of $x \mapsto x$ and of constant [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) is immediate from the definition, hence polynomials and rational [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) are [continuous](#def-g12-limcont-continuity). [Continuity](#def-g12-limcont-continuity) of the remaining usual [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) is proved in the chapters where they are constructed, or admitted. ∎

**Example 21.11.**

The *floor [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function)* $x \mapsto \floor{x}$ (the greatest [integer](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) $\leq x$) is not [continuous](#def-g12-limcont-continuity) at any [integer](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) $n$: its left limit there is $n - 1$ while $\floor{n} = n$.

![The floor function jumps at each integer: the value (filled dot) differs from the left limit (open dot).](https://one-course.com/images/onecourse/chapters/math-2/g12-limcont/fig-436d9712d64c.svg)

*The floor [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) jumps at each [integer](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets): the value (filled dot) differs from the left limit (open dot).*

**Proposition 21.12 (Sequences and continuity).**

If $u_n \to \ell$ and $f$ is [continuous](#def-g12-limcont-continuity) at $\ell$, then $f(u_n) \to f(\ell)$. In particular, if a [sequence](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) defined by $u_{n+1} = f(u_n)$ [converges](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-limit) to $\ell$ and $f$ is [continuous](#def-g12-limcont-continuity) at $\ell$, then $f(\ell) = \ell$.

**Proof.** The first claim is [Theorem 21.6](#thm-g12-limcont-composition) applied with the [sequence](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) $(u_n)$. For the second, pass to the limit on both sides of $u_{n+1} = f(u_n)$: the left side tends to $\ell$, the right side to $f(\ell)$, and limits are unique. ∎

## 21.3 The intermediate value theorem

**Theorem 21.13 (Intermediate value theorem).**

Let $f$ be [continuous](#def-g12-limcont-continuity) on $\intcc{a}{b}$ and let $k$ be any number between $f(a)$ and $f(b)$. Then there exists $c \in \intcc{a}{b}$ such that $f(c) = k$.

![The intermediate value theorem: a continuous curve joining (a, f(a)) to (b, f(b)) must cross every horizontal line y = k lying between f(a) and f(b).](https://one-course.com/images/onecourse/chapters/math-2/g12-limcont/fig-04c34a00da51.svg)

*The intermediate value theorem: a [continuous](#def-g12-limcont-continuity) curve joining $(a, f(a))$ to $(b, f(b))$ must cross every horizontal line $y = k$ lying between $f(a)$ and $f(b)$.*

**Proof.** Replacing $f$ by $f - k$ (and by $k - f$ if needed), we may assume $f(a) \leq 0 \leq f(b)$ and look for a zero of $f$. We construct two [sequences](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) by *dichotomy*: set $a_0 = a$, $b_0 = b$; given $a_n \leq b_n$ with $f(a_n) \leq 0 \leq f(b_n)$, let $m = \frac{a_n + b_n}{2}$ and define

$$
(a_{n+1}, b_{n+1}) =
\begin{cases}
(a_n, m) & \text{if } f(m) \geq 0,\\
(m, b_n) & \text{if } f(m) < 0.
\end{cases}
$$

In both cases $f(a_{n+1}) \leq 0 \leq f(b_{n+1})$, the [sequence](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) $(a_n)$ is [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic), $(b_n)$ is [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations), and $b_n - a_n = \frac{b-a}{2^n} \to 0$: the [sequences](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-sequence) are adjacent (see [Exercise 20.10](https://one-course.com/books/math/2/en/chapter/20-sequences#exo-g12-seq-10)) and converge to a common limit $c \in \intcc{a}{b}$. By [continuity](#def-g12-limcont-continuity), $f(a_n) \to f(c)$ and $f(b_n) \to f(c)$; passing to the limit in $f(a_n) \leq 0$ and $f(b_n) \geq 0$ gives $f(c) \leq 0 \leq f(c)$, so $f(c) = 0$. ∎

**Remark 21.14.**

The [dichotomy](#thm-g12-limcont-ivt) proof is an algorithm: it yields arbitrarily precise [approximations](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-approx) of a solution, halving the error at each step. This is how a computer can solve $f(x) = 0$ knowing only how to evaluate $f$.

**Theorem 21.15 (Bijection theorem).**

Let $f$ be [continuous](#def-g12-limcont-continuity) and *strictly [monotonic](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic)* on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $I$. Then for every $k$ strictly between the limits of $f$ at the endpoints of $I$, the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $f(x) = k$ has a *unique* solution in $I$.

**Proof.** Existence: choose $a, b \in I$ with $k$ between $f(a)$ and $f(b)$ (possible since the values of $f$ approach its endpoint limits), and apply the intermediate value theorem on $\intcc{a}{b}$. Uniqueness: if $x < y$, strict monotonicity gives $f(x) \neq f(y)$, so two distinct points cannot both be solutions. ∎

**Method 21.16 (Solving f(x)=kf(x) = kf(x)=k with the bijection theorem).**

To prove that $f(x) = k$ has exactly one solution in an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval):

1. establish a [variation table](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-table) of $f$ (sign of $f'$ , see [Chapter 22](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#ch-g12-deriv) );
2. on each [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) where $f$ is [continuous](#def-g12-limcont-continuity) and strictly [monotonic](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic) , compare $k$ with the endpoint values or limits;
3. apply the bijection theorem on each such [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) and count the solutions.

A calculator or [dichotomy](#thm-g12-limcont-ivt) then localizes each solution as finely as desired.

**Example 21.17.**

The [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $x^3 + x = 1$ has a unique real solution. Indeed $f(x) = x^3 + x$ is [continuous](#def-g12-limcont-continuity) and strictly [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic) on $\R$ (as a sum of strictly [increasing functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-monotone)), $\lim_{-\infty} f = -\infty$ and $\lim_{+\infty} f = +\infty$, so the bijection theorem applies with $k = 1$. Since $f(0.6) = 0.816 < 1$ and $f(0.7) = 1.043 > 1$, the solution lies in $\intoo{0.6}{0.7}$.

## 21.4 Exercises

**Exercise 21.1 ★.**

Compute the following limits:

$$
\lim_{x\to+\infty} \frac{2x^3 - x + 1}{5x^3 + x^2}, \qquad
\lim_{x\to-\infty} \bigl(x^2 + 3x - 1\bigr), \qquad
\lim_{x\to 2^+} \frac{x+1}{x-2}, \qquad
\lim_{x\to+\infty} \frac{\cos x}{x}.
$$

**Solution of Exercise 21.1.**

Dividing by $x^3$: $\dfrac{2x^3 - x + 1}{5x^3 + x^2} = \dfrac{2 - 1/x^2 + 1/x^3}{5 + 1/x}
\to \dfrac{2}{5}$.

$x^2 + 3x - 1 = x^2(1 + 3/x - 1/x^2) \to +\infty$ as $x \to -\infty$, since $x^2 \to +\infty$ and the bracket tends to $1$.

As $x \to 2^+$, the numerator tends to $3 > 0$ and the denominator to $0^+$, so $\dfrac{x+1}{x-2} \to +\infty$.

For $x > 0$, $-\dfrac1x \leq \dfrac{\cos x}{x} \leq \dfrac1x$, so the limit is $0$ by the squeeze theorem.

**Exercise 21.2 ★.**

Let $f(x) = \dfrac{2x^2 - 3x + 1}{x - 1}$, defined for $x \neq 1$.

1. Show that $f(x) = 2x - 1$ for all $x \neq 1$ .
2. Deduce $\lim\limits_{x \to 1} f(x)$ . Can $f$ be extended to a [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) [continuous](#def-g12-limcont-continuity) at $1$ ?

**Solution of Exercise 21.2.**

*1.* $2x^2 - 3x + 1 = (x-1)(2x-1)$ (the [root](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) $x=1$ is apparent from $2 - 3 + 1 = 0$). For $x \neq 1$, cancel $x - 1$: $f(x) = 2x - 1$.

*2.* Hence $\lim_{x\to1} f(x) = 1$. Setting $f(1) = 1$ extends $f$ to the polynomial [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $x \mapsto 2x-1$ on $\R$, which is [continuous](#def-g12-limcont-continuity) at $1$.

**Exercise 21.3 ★.**

Determine the [asymptotes](#def-g12-limcont-asymptote) of the curve of $f(x) = \dfrac{3x - 1}{x + 2}$ and sketch the curve.

**Solution of Exercise 21.3.**

Write $f(x) = \dfrac{3x - 1}{x + 2} = \dfrac{3(x+2) - 7}{x+2}
= 3 - \dfrac{7}{x+2}$. As $x \to \pm\infty$, $f(x) \to 3$: the line $y = 3$ is a [horizontal asymptote](#def-g12-limcont-asymptote) at both infinities. As $x \to (-2)^\pm$, $f(x) \to \mp\infty$: the line $x = -2$ is a [vertical asymptote](#def-g12-limcont-asymptote). The curve is a [hyperbola](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-reference), below $y=3$ for $x > -2$ and above it for $x < -2$.

**Exercise 21.4 ★★.**

Compute

$$
\lim_{x\to+\infty} \bigl(\sqrt{x^2 + x} - x\bigr)
\qquad\text{and}\qquad
\lim_{x\to 0} \frac{\sqrt{1+x} - 1}{x}.
$$

**Solution of Exercise 21.4.**

Multiply by the conjugate:

$$
\sqrt{x^2 + x} - x = \frac{x^2 + x - x^2}{\sqrt{x^2+x} + x}
= \frac{x}{x\left(\sqrt{1 + 1/x} + 1\right)}
= \frac{1}{\sqrt{1 + 1/x} + 1}
\xrightarrow[x\to+\infty]{} \frac{1}{2}.
$$

Similarly,

$$
\frac{\sqrt{1+x} - 1}{x} = \frac{(1+x) - 1}{x\left(\sqrt{1+x} + 1\right)}
= \frac{1}{\sqrt{1+x} + 1} \xrightarrow[x\to0]{} \frac{1}{2}.
$$

**Exercise 21.5 ★★.**

Let $f(x) = x + \sin x$.

1. Show that $f(x) \to +\infty$ as $x \to +\infty$ , although $f$ is not [monotonic](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic) on any [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $\intco{A}{+\infty}$ .
2. Show that $g(x) = \dfrac{x + \sin x}{x - \sin x}$ tends to $1$ at $+\infty$ .

**Solution of Exercise 21.5.**

*1.* Since $\sin x \geq -1$, $f(x) \geq x - 1 \to +\infty$, and we conclude by comparison. Yet $f'(x) = 1 + \cos x$ vanishes at every odd multiple of $\pi$, and on every [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $\intco{A}{+\infty}$ the [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) alternates between [intervals](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) of strict increase; it is [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic) but not strictly, and more importantly its growth is not needed: comparison suffices.

*2.* For $x > 1$, divide by $x$:

$$
g(x) = \frac{1 + \frac{\sin x}{x}}{1 - \frac{\sin x}{x}},
$$

and $\frac{\sin x}{x} \to 0$ by the squeeze theorem ($\abs{\sin x} \leq 1$). Hence $g(x) \to \frac{1+0}{1-0} = 1$.

**Exercise 21.6 ★★.**

Show that the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $x^3 - 3x + 1 = 0$ has exactly three real solutions, and give for each an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) of length $1$ containing it. (Hint: study the variations of $f(x) = x^3 - 3x + 1$; its [derivative](https://one-course.com/books/math/2/en/chapter/12-differentiation-a-first-course#def-g11-deriv-derivative) is $3x^2 - 3$.)

**Solution of Exercise 21.6.**

$f(x) = x^3 - 3x + 1$ is [continuous](#def-g12-limcont-continuity), with $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$: $f$ increases on $\intoc{-\infty}{-1}$, decreases on $\intcc{-1}{1}$, increases on $\intco{1}{+\infty}$, with local [maximum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $f(-1) = 3 > 0$ and local [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $f(1) = -1 < 0$, $\lim_{-\infty} f = -\infty$, $\lim_{+\infty} f = +\infty$.

On each of the three monotonicity [intervals](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval), $f$ is [continuous](#def-g12-limcont-continuity), strictly [monotonic](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic), and $0$ lies strictly between the endpoint values/limits, so the bijection theorem gives exactly one zero per [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval): three zeros in total. Sign checks: $f(-2) = -1 < 0 < f(-1) = 3$, so one zero in $\intoo{-2}{-1}$; $f(0) = 1 > 0 > f(1) = -1$, one zero in $\intoo{0}{1}$; $f(1) = -1 < 0 < f(2) = 3$, one zero in $\intoo{1}{2}$.

**Exercise 21.7 ★★.**

Let $f \colon \intcc{0}{1} \to \intcc{0}{1}$ be [continuous](#def-g12-limcont-continuity). Show that $f$ has a *[fixed point](https://one-course.com/books/math/2/en/chapter/3-functions#pb-g10-functions-1)*: there exists $c \in \intcc{0}{1}$ with $f(c) = c$. (Hint: apply the intermediate value theorem to $g(x) = f(x) - x$.)

**Solution of Exercise 21.7.**

Let $g(x) = f(x) - x$, [continuous](#def-g12-limcont-continuity) on $\intcc{0}{1}$. Since $f(0) \in \intcc{0}{1}$, $g(0) = f(0) \geq 0$; since $f(1) \in \intcc{0}{1}$, $g(1) = f(1) - 1 \leq 0$. By the intermediate value theorem, $g$ vanishes at some $c \in \intcc{0}{1}$, i.e. $f(c) = c$.

**Exercise 21.8 ★★★.**

A hiker climbs a mountain trail, starting at 8:00 and arriving at 20:00. The next day she descends the same trail, again starting at 8:00 and arriving at 20:00. Show that there is a time of day at which she is at exactly the same point of the trail on both days. (Hint: introduce the difference of the two position [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function).)

**Solution of Exercise 21.8.**

Let $u(t)$ and $d(t)$ be the positions (measured as distance from the bottom of the trail) at time $t \in \intcc{8}{20}$ on the way up and on the way down. Both are [continuous](#def-g12-limcont-continuity), $u(8) = 0$, $u(20) = L$ (the length of the trail), $d(8) = L$, $d(20) = 0$. The [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $g = u - d$ is [continuous](#def-g12-limcont-continuity) with $g(8) = -L \leq 0$ and $g(20) = L \geq 0$, so $g(c) = 0$ for some time $c$ by the intermediate value theorem: at time $c$, both days’ positions coincide.

**Exercise 21.9 ★★★.**

Let $f(x) = x^5 + x^3 + x - 1$.

1. Show that $f$ has a unique real zero $\alpha$ , and that $\alpha \in \intoo{0}{1}$ .
2. Describe a [dichotomy](#thm-g12-limcont-ivt) procedure computing $\alpha$ to within $10^{-6}$ , and give the number of iterations required starting from $\intcc{0}{1}$ .

**Solution of Exercise 21.9.**

*1.* $f'(x) = 5x^4 + 3x^2 + 1 > 0$, so $f$ is strictly [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic) on $\R$; with $\lim_{\pm\infty} f = \pm\infty$ and [continuity](#def-g12-limcont-continuity), the bijection theorem gives a unique real zero $\alpha$. Since $f(0) = -1 < 0$ and $f(1) = 2 > 0$, $\alpha \in \intoo{0}{1}$.

*2.* Maintain an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $\intcc{a_n}{b_n}$ with $f(a_n) < 0 < f(b_n)$, starting from $\intcc{0}{1}$; at each step evaluate $f$ at the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) $m$ and keep the [half-interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) on which the sign change persists. After $n$ steps the [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) has length $2^{-n}$, so the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) approximates $\alpha$ to within $2^{-(n+1)}$. We need $2^{-(n+1)} \leq 10^{-6}$, i.e. $n + 1 \geq \frac{6}{\log_{10} 2} \approx 19.93$: $n = 20$ iterations suffice.

## 21.5 Problem: You cannot cross a river without getting wet

**Problem 21.1.**

Weekend problem — the intermediate value theorem forces fixed points, equal-temperature antipodes and steady tables: three impossible-sounding theorems from one idea

Crumple a map of France and drop it anywhere on French soil: some point of the map lies *exactly* above the place it represents. At every instant, two diametrically opposite points of the equator have *exactly* the same temperature. And a wobbly square table on bumpy ground can always be steadied by *rotating* it. Three party tricks, one theorem: a [continuous](#def-g12-limcont-continuity) quantity cannot pass from negative to positive without vanishing ([Theorem 21.13](#thm-g12-limcont-ivt)). This problem trains limits, then builds all three wonders — and knows exactly what the theorem does not promise.

**Part I — Limit fluency.**

1. Compute $\lim_{x \to +\infty} \dfrac{3x^2 - x}{x^2 + 5}$ , $\lim_{x \to +\infty} \dfrac{2x + 1}{x^2 + 1}$ , and $\lim_{x \to +\infty} (x^3 - x^2)$ ( [Method 21.8](#met-g12-limcont-limits) ).
2. Give the [asymptotes](#def-g12-limcont-asymptote) of $f(x) = \dfrac{3x - 1}{x + 2}$ ( [Definition 21.4](#def-g12-limcont-asymptote) ), computing the relevant limits.
3. Compute $\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$ (factor) and $\lim_{x \to 0} \dfrac{\sqrt{x + 1} - 1}{x}$ (conjugate).
4. By the squeeze theorem ( [Theorem 21.7](#thm-g12-limcont-squeeze) ): compute $\lim_{x \to 0} x \sin\frac1x$ and $\lim_{x \to +\infty} \frac{\sin x}{x}$ .
5. By composition and dominant terms: $\lim_{x \to +\infty} \dfrac{\sqrt{4x^2 + x}}{x}$ .

**Part II — [Roots](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant), certified.**

6. Show that the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $x^3 + x = 5$ has at least one solution in $\intcc{1}{2}$ .
7. Show the solution is unique on $\R$ ( [Theorem 21.15](#thm-g12-limcont-bijection) ).
8. Prove the classic: *every polynomial of odd degree has at least one real [root](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant)* . (Write out the argument for $p(x) = x^3 + a x^2 + b x + c$ : compute the two infinite limits by dominant term, then apply the IVT on a large [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) .)
9. Locating the [root](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) of question 6 by [dichotomy](#thm-g12-limcont-ivt) : how many halvings of $\intcc{1}{2}$ guarantee an error below $10^{-6}$ ( [Exercise 21.9](#exo-g12-limcont-9) )? Compare with the digit-doubling of Newton’s method ( [Problem 12.1](https://one-course.com/books/math/2/en/chapter/12-differentiation-a-first-course#pb-g11-deriv-1) ): roughly how many steps would it need?
10. The counterexample that guards the theorem: $g(x) = \frac1x$ satisfies $g(-1) = -1 < 0$ and $g(1) = 1 > 0$ , yet never vanishes. Which hypothesis of the IVT fails — and what does the example teach about checking hypotheses?

**Part III — Three impossible-sounding theorems.**

11. The fixed-point theorem of the segment: if $f : \intcc{0}{1} \to \intcc{0}{1}$ is [continuous](#def-g12-limcont-continuity) , prove that some $c$ satisfies $f(c) = c$ . (Study $g(x) = f(x) - x$ at the two ends.)
12. Translate question 11 into the crumpled map: what plays the role of $f$ , why does it land in $\intcc{0}{1}$ (figuratively: within France), and what is the point lying exactly above its own location? (One dimension is admitted silently — the honest two-dimensional statement is Brouwer’s theorem, university volumes.)
13. The equator: let $T$ be a [continuous](#def-g12-limcont-continuity) temperature on a circle, seen as a $2\pi$ -periodic [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) with $T(0) = T(2\pi)$ . Set $g(t) = T(t) - T(t + \pi)$ . Compute $g(0) + g(\pi)$ , deduce that $g$ vanishes somewhere on $\intcc{0}{\pi}$ , and state the theorem about antipodal points.
14. The wobbly table: a perfectly rigid square table stands on [continuous](#def-g12-limcont-continuity) (no steps!) bumpy ground; three legs always touch, one hovers by a height $h(\theta)$ that depends [continuously](#def-g12-limcont-continuity) on the table’s rotation angle $\theta$ . Explain why rotating by a quarter turn forces the hovering gap to change sign — and conclude that some angle $\theta^*$ steadies all four legs. (This argument was published as a genuine theorem in 2005; café owners knew the trick first.)
15. The hiker of [Exercise 21.8](#exo-g12-limcont-8) used the same scheme. State the template shared by questions 11, 13, 14 and the hiker — build which [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) , check what at the ends, invoke what — in three lines.
16. What the IVT does *not* give: existence came free, but which two questions does it leave open, and which tools of this problem answer them (questions 7 and 9)?

**Part IV — Portraits with [asymptotes](#def-g12-limcont-asymptote).**

17. Rewrite $f(x) = \dfrac{3x - 1}{x + 2}$ as $3 - \dfrac{7}{x + 2}$ and use it to describe the curve completely: [asymptotes](#def-g12-limcont-asymptote) , variations on each side, sketch. (A shifted [hyperbola](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-reference) — the average-cost curve of [Problem 4.1](https://one-course.com/books/math/2/en/chapter/4-reference-functions#pb-g10-reffunc-1) was one too.)
18. The [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $f(x) = \dfrac{x^2 - 1}{x - 1}$ is undefined at $1$ . Which value assigned to $f(1)$ makes it [continuous](#def-g12-limcont-continuity) there — and why do analysts call such a point a *removable* discontinuity?
19. The floor [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) $x \mapsto \lfloor x \rfloor$ (greatest [integer](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) $\leq x$ ): where is it [continuous](#def-g12-limcont-continuity) , where does it jump, and why does the IVT’s conclusion fail for it (find $a < b$ with $\lfloor a \rfloor < \frac12 < \lfloor b \rfloor$ yet no solution of $\lfloor x \rfloor = \frac12$ )?
20. Finale — [continuity](#def-g12-limcont-continuity) ’s two faces: the local promise (value $=$ limit: no teleporting) and the global power (IVT: every intermediate value visited). File the four wonders of Part III under the right face, and give the river-crossing slogan its precise mathematical name.

**Solution of Problem 21.1.**

**1.** Dominant terms: $\frac{3x^2}{x^2} \to 3$; $\frac{2x}{x^2} = \frac2x \to 0$; $x^3 - x^2 = x^3\left(1 - \frac1x\right) \to +\infty$.

**2.** $\lim_{x \to \pm\infty} f = 3$: [horizontal asymptote](#def-g12-limcont-asymptote) $y = 3$; $\lim_{x \to (-2)^\pm} f = \mp\infty$: [vertical asymptote](#def-g12-limcont-asymptote) $x = -2$.

**3.** $\frac{(x-3)(x+3)}{x - 3} = x + 3 \to 6$. And $\frac{\sqrt{x+1} - 1}{x} = \frac{x}{x(\sqrt{x+1} + 1)} =
\frac{1}{\sqrt{x+1} + 1} \to \frac12$.

**4.** $\abs{x \sin\frac1x} \leq \abs x \to 0$, so the limit is $0$. And $\abs{\frac{\sin x}{x}} \leq \frac1x \to 0$: limit $0$.

**5.** $\frac{\sqrt{4x^2 + x}}{x} =
\sqrt{4 + \frac1x} \to \sqrt4 = 2$ ([continuity](#def-g12-limcont-continuity) of the square root, composition).

**6.** $f(x) = x^3 + x$ is [continuous](#def-g12-limcont-continuity) with $f(1) = 2 < 5 < 10 = f(2)$: by the IVT, $f(c) = 5$ for some $c \in \intoo{1}{2}$.

**7.** $f'(x) = 3x^2 + 1 > 0$: strictly [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic) on $\R$, [continuous](#def-g12-limcont-continuity), with limits $\mp\infty$: by the bijection theorem, every value — in particular $5$ — is attained exactly once.

**8.** $p(x) = x^3\left(1 + \frac ax + \frac{b}{x^2} +
\frac{c}{x^3}\right)$: as $x \to +\infty$, $p \to +\infty$; as $x \to -\infty$, $p \to -\infty$. So on some huge [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $\intcc{-M}{M}$, $p(-M) < 0 < p(M)$: the IVT delivers a [root](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant). The same dominant-term argument works verbatim for any odd degree — odd powers keep opposite signs at the two infinities.

**9.** Halving $n$ times leaves an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) of length $2^{-n}$; $2^{-20} \approx 9.5 \times 10^{-7} < 10^{-6}$: twenty steps. Newton doubles correct digits each step: from one correct digit, about $3$–$4$ steps reach six — the tortoise and the greyhound of root-finding.

**10.** $g$ is not [continuous](#def-g12-limcont-continuity) on $\intcc{-1}{1}$: it explodes at $0$, which the [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) contains. The IVT’s conclusion evaporates with its hypothesis — always check [continuity](#def-g12-limcont-continuity) *on the whole [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval)*, not just at the endpoints.

**11.** $g(x) = f(x) - x$ is [continuous](#def-g12-limcont-continuity); $g(0) = f(0) \geq 0$ (values lie in $\intcc{0}{1}$) and $g(1) = f(1) - 1 \leq 0$. By the IVT, $g(c) = 0$ for some $c$: $f(c) = c$.

**12.** Idealizing the country as the segment $\intcc{0}{1}$: $f$ sends each place to the place shown directly beneath it on the crumpled map; crumpling keeps the map over the country, so $f$ lands in $\intcc{0}{1}$, and it is [continuous](#def-g12-limcont-continuity) (paper does not tear). The [fixed point](https://one-course.com/books/math/2/en/chapter/3-functions#pb-g10-functions-1) $c$ is the spot lying precisely above its own [image](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) — Brouwer’s theorem handles the honest two-dimensional map.

**13.** $g(0) + g(\pi) = \left(T(0) - T(\pi)\right) +
\left(T(\pi) - T(2\pi)\right) = T(0) - T(2\pi) = 0$. So $g(0)$ and $g(\pi)$ are opposite (or both zero): either way $g$, [continuous](#def-g12-limcont-continuity), vanishes at some $t^* \in \intcc{0}{\pi}$: $T(t^*) = T(t^* + \pi)$ — somewhere on the equator, two antipodal points agree on the temperature, at every single moment.

**14.** Rotating the table by a quarter turn exchanges the roles of the two diagonals: the pair of legs that pressed the ground and the pair that (one of them) hovered swap situations, so the hovering height — counted with its sign as “gap of the distinguished diagonal” — takes opposite signs at $\theta$ and $\theta + \frac\pi2$. [Continuous](#def-g12-limcont-continuity) ground makes $h$ [continuous](#def-g12-limcont-continuity), and the IVT hands over an angle $\theta^*$ with $h(\theta^*) = 0$: four legs down. (No such luck with a rectangular table or stepped ground.)

**15.** Template: build the *difference* of the two quantities you want to coincide ($f(x) - x$; $T(t) - T(t + \pi)$; up-position minus down-position for the hiker; the signed gap for the table); check it takes values of both signs (often via a symmetry of the ends); invoke the IVT on a [continuous](#def-g12-limcont-continuity) [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function). Coincidence forced.

**16.** Open questions: *how many* solutions (answered by strict monotonicity and the bijection theorem, question 7) and *where* they are (answered by [dichotomy](#thm-g12-limcont-ivt) or Newton, question 9). The IVT is a pure existence oracle.

**17.** $3 - \frac{7}{x + 2}$: [vertical asymptote](#def-g12-limcont-asymptote) $x = -2$, horizontal $y = 3$; on each side of $-2$ the [function](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) is [increasing](https://one-course.com/books/math/2/en/chapter/20-sequences#def-g12-seq-monotonic) ($-\frac{7}{x+2}$ increases where defined); the curve is the reference [hyperbola](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-reference) $-\frac7x$ shifted left $2$ and up $3$ — two branches hugging the [asymptote](#def-g12-limcont-asymptote) cross.

**18.** For $x \neq 1$: $f(x) = x + 1$, so $\lim_{x \to 1} f = 2$: setting $f(1) = 2$ restores [continuity](#def-g12-limcont-continuity). The discontinuity was a puncture, not a jump — fill the single missing point and the curve heals: *removable*.

**19.** [Continuous](#def-g12-limcont-continuity) on every [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) between consecutive [integers](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets); at each [integer](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) $n$ it jumps from $n - 1$ to $n$. With $a = 0.9$, $b = 1.1$: $\lfloor a \rfloor = 0 < \frac12 <
1 = \lfloor b \rfloor$, yet $\lfloor x \rfloor = \frac12$ has no solution — the jump teleports over $\frac12$, which is precisely what the IVT forbids [continuous](#def-g12-limcont-continuity) [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) to do.

**20.** Local face: value $=$ limit at each point — the removable puncture (question 18) violated it curably, the floor’s jumps (question 19) incurably. Global face: on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval), all intermediate values are visited — the [root](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) certificates (6–8), the [fixed point](https://one-course.com/books/math/2/en/chapter/3-functions#pb-g10-functions-1), the equator, the table and the hiker all live here. The slogan’s official name: the *intermediate value theorem*.