Chapter 5 · Continuity and Differentiability

Example 16: Every rational function $\frac{p(x)}{q(x)}$ is continuous

ExamplesAlgebra of continuous functions · Section 5.2.1

The question

Prove that every rational function is continuous.

Exam answer

Let $f(x) = \dfrac{p(x)}{q(x)}$ ⁠, where $p$ and $q$ are polynomials. Its domain is all real $x$ with $q(x) \ne 0$ ⁠.

Let $c$ be in the domain, so $q(c) \ne 0$ ⁠. Since polynomials are continuous, $\lim_{x\to c} p(x) = p(c)$ and $\lim_{x\to c} q(x) = q(c)$ ⁠. So

\begin{aligned} \lim_{x\to c} f(x) &= \frac{\lim_{x\to c} p(x)}{\lim_{x\to c} q(x)} \\ &= \frac{p(c)}{q(c)} = f(c) \end{aligned}

$\therefore\ f$ is continuous at every point of its domain, so every rational function is continuous. Hence proved.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A rational function is a quotient of two polynomials, $\frac{p(x)}{q(x)}$ ⁠. Polynomials are continuous everywhere (Example 14), and Theorem 1(4) says a quotient of two functions continuous at $c$ is continuous at $c$ provided the denominator is not 0 at $c$ . That proviso is exactly the condition for $c$ to be in the domain of the rational function. So at every point of its domain the rational function is continuous.

Given, to show, and the plan

Given: $f(x) = \frac{p(x)}{q(x)}$ with $p$ ⁠, $q$ polynomial functions, defined where $q(x) \ne 0$ ⁠.
To show: $f$ is continuous at every point of its domain, so $f$ is a continuous function.
Plan: Identify the domain. Fix a point $c$ of the domain. Use Example 14 for $p$ and $q$ ⁠, then Theorem 1(4) with $q(c) \ne 0$ ⁠. Finally apply Definition 2, and explain the zeros of $q$ ⁠.

Basics you need

1Rational function and its domain

$f(x) = \frac{p(x)}{q(x)}$ ⁠, $p$ and $q$ polynomials, $q$ not the zero polynomial. Division by 0 is not allowed, so the domain is $\{x \in \R : q(x) \ne 0\}$ ⁠. Example: $\frac{x+1}{x^2-4}$ has domain $\R$ without $-2$ and $2$ ⁠.

2Every polynomial is continuous

For every polynomial $p$ and every real $c$ ⁠, $\lim_{x\to c} p(x) = p(c)$ (Example 14).

3Theorem 1(4)

If $f$ and $g$ are continuous at $c$ and $g(c) \ne 0$ ⁠, then $\frac{f}{g}$ is continuous at $c$ ⁠. It rests on the quotient rule of limits, which needs the limit of the denominator to be non-zero. (The book proves only part (1) of Theorem 1 and leaves part (4) to the reader; Step 3 below carries out that argument for $\frac{p}{q}$ ⁠.) Lesson: algebra of continuous functions

4Definition 2

A function is continuous if it is continuous at every point of its domain; points outside the domain are not tested. Lesson: continuity at a point

Solution

The domain of $f$

$f(x) = \frac{p(x)}{q(x)}$ can be computed exactly when the denominator is non-zero. So
$\text{domain of } f = \{x \in \R : q(x) \ne 0\}.$
For example, $\frac{x+1}{x^2-4}$ is defined at every real number except $x = -2$ and $x = 2$ ⁠, where $x^2 - 4 = 0$ ⁠.

We need the domain first because Definition 2 asks for continuity at every point of the domain, and only there. A non-zero polynomial has finitely many zeros, so the domain is all of $\R$ except finitely many points (possibly none, as for $\frac{1}{x^2+1}$ ⁠).

The book's notation " $f(x) = \frac{p(x)}{q(x)},\ q(x) \ne 0$ " records the domain in the formula itself.
Fix a point $c$ of the domain

Let $c$ be any real number with $q(c) \ne 0$ ⁠. By Example 14, both polynomials are continuous at $c$ ⁠:
$\lim_{x\to c} p(x) = p(c), \qquad \lim_{x\to c} q(x) = q(c).$
Polynomials are continuous at every real number, so in particular at our $c$ ⁠. These two limits are the two ingredients Theorem 1(4) needs.

Example 14: every polynomial function is continuous.
Apply Theorem 1(4)

$p$ and $q$ are continuous at $c$ and $q(c) \ne 0$ ⁠, so by Theorem 1(4) the quotient $\frac{p}{q} = f$ is continuous at $c$ ⁠. In terms of limits:
$\begin{aligned} \lim_{x\to c} \frac{p(x)}{q(x)} &= \frac{\lim_{x\to c} p(x)}{\lim_{x\to c} q(x)} \\ &= \frac{p(c)}{q(c)} = f(c). \end{aligned}$
The quotient rule of limits ("the limit of a quotient is the quotient of the limits") only works when the limit of the denominator is not 0; otherwise the right-hand side would be a division by 0. Here that limit is $q(c)$ ⁠, which is non-zero precisely because $c$ is in the domain. This is the "provided $g(c) \ne 0$ " in Theorem 1(4).

Theorem 1(4), Section 5.2.1. Lesson: algebra of continuous functions
Conclude with Definition 2

$c$ was an arbitrary point of the domain, so $f$ is continuous at every point of its domain. By Definition 2, every rational function is continuous.

This is the textbook's proof: "since polynomial functions are continuous (Example 14), $f$ is continuous by (4) of Theorem 1". Steps 1 to 3 spell out what each part of that sentence contributes.
What happens at the zeros of $q$ ⁠?

At a zero $a$ of $q$ ⁠, the number $f(a)$ does not exist, so $a$ is not in the domain, and in this book's sense $a$ is not a point of discontinuity. The graph may break there (for $\frac{x+1}{x^2-4}$ ⁠, it shoots off to $\pm\infty$ near $x = \pm 2$ ⁠), but this does not stop $f$ being a continuous function.

This is the same situation as $\frac{1}{x}$ in Example 9 and the gap at 0 in Example 12: continuity is a property of points in the domain, so a missing point is not a failure.

Do not cancel factors before finding the domain. $\frac{x^2 - 1}{x - 1}$ equals $x + 1$ for $x \ne 1$ ⁠, but it is not defined at $x = 1$ ⁠, so its domain is $\R$ without 1. Lesson: continuity at a point

Result

A rational function $f = \frac{p}{q}$ has domain $\{x : q(x) \ne 0\}$ ⁠, and at each point $c$ of it, $\lim_{x\to c} f(x)$ equals $\frac{p(c)}{q(c)} = f(c)$ ⁠. Hence every rational function is continuous (at every point of its domain). This agrees with the textbook's solution.

Check it with numbers

Take $f(x) = \frac{x+1}{x^2-4}$ and $c = 1$ ⁠: $f(1) = \frac{2}{-3} \approx -0.666667$ ⁠. Nearby,

f(0.999) = \tfrac{1.999}{-3.001999} \approx -0.665890,

f(1.001) = \tfrac{2.001}{-2.997999} \approx -0.667445,

both within $0.001$ of $f(1)$ ⁠, as continuity at 1 predicts.

Common mistakes

Calling the zeros of $q$ points of discontinuity. They are not in the domain, so in this book's sense they are neither points of continuity nor of discontinuity.
Dropping the condition $q(c) \ne 0$ ⁠. Theorem 1(4) needs it; without it the quotient rule of limits would divide by 0.
Simplifying first. Cancelling $x - 1$ in $\frac{x^2-1}{x-1}$ gives a different function (one defined at 1).

What to take away

Rational functions are continuous wherever they are defined, that is, wherever the denominator is non-zero.
Theorem 1 turns known continuous functions into new ones; here Example 14 plus Theorem 1(4) gives the whole result in one line.
The same pattern proves $\tan x = \frac{\sin x}{\cos x}$ continuous in Example 18.

Every fact used, and where it comes from

Fact	Used in	Source
Rational function $\frac{p(x)}{q(x)}$ ⁠, $q(x) \ne 0$	Steps 1, 5	Example 16, Section 5.2.1; Class XI
Every polynomial function is continuous	Step 2	Example 14, Section 5.2
$\frac{f}{g}$ is continuous at $c$ if $f$ ⁠, $g$ are and $g(c) \ne 0$	Step 3	Theorem 1(4), Section 5.2.1 (Lesson: algebra of continuous functions)
Quotient rule of limits	Step 3	Algebra of limits, Class XI (Basics: limits)
Continuous function; points outside the domain are not tested	Steps 4, 5	Definitions 1 and 2, Section 5.2 (Lesson: continuity at a point)

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Example 16: Every rational function p(x)q(x)\frac{p(x)}{q(x)}q(x)p(x)​ is continuous

Basics you need

Solution

The domain of fff

Fix a point ccc of the domain

Apply Theorem 1(4)

Conclude with Definition 2

What happens at the zeros of qqq⁠?

Every fact used, and where it comes from

Example 16: Every rational function $\frac{p(x)}{q(x)}$ is continuous

The domain of $f$

Fix a point $c$ of the domain

What happens at the zeros of $q$ ⁠?