Chapter 5 · Continuity and Differentiability

Example 18: $\tan x$ is a continuous function

ExamplesAlgebra of continuous functions · Section 5.2.1

The question

Prove that the function defined by $f(x) = \tan x$ is a continuous function.

Exam answer

$f(x) = \tan x = \dfrac{\sin x}{\cos x}$ is defined when $\cos x \ne 0$ ⁠, that is, for $x \ne (2n + 1)\frac{\pi}{2}$ ⁠, $n \in \Z$ ⁠.

Let $c$ be in the domain, so $\cos c \ne 0$ ⁠. Since $\sin x$ and $\cos x$ are continuous,

\begin{aligned} \lim_{x\to c}\tan x &= \frac{\lim_{x\to c}\sin x}{\lim_{x\to c}\cos x} \\ &= \frac{\sin c}{\cos c} = \tan c = f(c) \end{aligned}

$\therefore\ f$ is continuous at every point of its domain, so $f(x) = \tan x$ is a continuous function. Hence proved.

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

The one idea this problem is built on

$\tan x = \frac{\sin x}{\cos x}$ is a quotient of two continuous functions (Example 17 and its Remark). Theorem 1(4) makes a quotient continuous at every point where the denominator is not 0, and " $\cos x \ne 0$ " is exactly the condition for $\tan x$ to be defined. So $\tan x$ is continuous at every point of its domain. The breaks in its graph occur only at the excluded points $x = (2n+1)\frac{\pi}{2}$ ⁠.

Given, to show, and the plan

Given: $f(x) = \tan x = \frac{\sin x}{\cos x}$ ⁠; $\sin x$ and $\cos x$ are continuous at every real number (Example 17).
To show: $f$ is continuous at every point of its domain, so it is a continuous function.
Plan: Find where $\cos x = 0$ to get the domain. At each point $c$ of the domain apply Theorem 1(4). Conclude with Definition 2, and explain the excluded points.

Basics you need

1Where $\cos x = 0$

$\cos x = 0$ exactly when $x$ is an odd multiple of $\frac{\pi}{2}$ ⁠: $x = (2n+1)\frac{\pi}{2}$ ⁠, $n \in \Z$ ⁠, that is $\dots, -\frac{3\pi}{2}, -\frac{\pi}{2},$ $\frac{\pi}{2}, \frac{3\pi}{2}, \dots$ (Class XI). Basics: trigonometric identities

2Sine and cosine are continuous

Proved in Example 17 (sine) and its Remark (cosine): $\lim_{x\to c}\sin x = \sin c$ and $\lim_{x\to c}\cos x = \cos c$ for every real $c$ ⁠.

3Theorem 1(4)

If $f$ and $g$ are continuous at $c$ and $g(c) \ne 0$ ⁠, then $\frac{f}{g}$ is continuous at $c$ ⁠. The book proves only part (1) and leaves part (4) to the reader; it follows from the quotient rule of limits (Class XI), as Step 3 shows for $\frac{\sin x}{\cos x}$ ⁠. Lesson: algebra of continuous functions

4Definition 2

A function is continuous if it is continuous at every point of its domain. Lesson: continuity at a point

Solution

Find the domain

$\tan x = \frac{\sin x}{\cos x}$ is defined exactly when $\cos x \ne 0$ ⁠. Since $\cos x = 0$ precisely at the odd multiples of $\frac{\pi}{2}$ ⁠,
$\text{domain} = \Big\{x \in \R : x \ne (2n+1)\tfrac{\pi}{2},\ n \in \Z\Big\}.$
Division by 0 is undefined, so every zero of the denominator must be removed. Knowing the domain tells us exactly which points need to be checked, as Definition 2 requires.

On the unit circle, $\cos x$ is the horizontal coordinate, which is 0 only at the top $(0, 1)$ and bottom $(0, -1)$ ⁠, that is at angles $\frac{\pi}{2} + n\pi = (2n+1)\frac{\pi}{2}$ ⁠. Basics: trigonometric identities
Fix a point $c$ of the domain

Let $c$ be any real number with $c \ne (2n+1)\frac{\pi}{2}$ ⁠, so $\cos c \ne 0$ ⁠. By Example 17 and its Remark,
$\lim_{x\to c}\sin x = \sin c, \qquad \lim_{x\to c}\cos x = \cos c.$
These two continuity facts hold at every real number, so in particular at our $c$ ⁠. They are the two ingredients needed by Theorem 1(4).

Example 17: sine and cosine are continuous.
Apply Theorem 1(4)

$\sin$ and $\cos$ are continuous at $c$ and $\cos c \ne 0$ ⁠, so $\frac{\sin}{\cos} = \tan$ is continuous at $c$ ⁠:
$\begin{aligned} \lim_{x\to c}\tan x &= \frac{\lim_{x\to c}\sin x}{\lim_{x\to c}\cos x} \\ &= \frac{\sin c}{\cos c} = \tan c. \end{aligned}$
The quotient of the limits makes sense because the limit of the denominator, $\cos c$ ⁠, is not 0. That is exactly the proviso " $g(c) \ne 0$ " of Theorem 1(4), and it holds because $c$ is in the domain.

Lesson: algebra of continuous functions
Conclude with Definition 2

$c$ was an arbitrary point of the domain, so $\tan x$ is continuous at every point of its domain. By Definition 2, $f(x) = \tan x$ is a continuous function.

This is the book's sentence " $\tan x$ being a quotient of two continuous functions is continuous wherever it is defined", with each part justified.
The excluded points

At $x = (2n+1)\frac{\pi}{2}$ the value $\tan x$ does not exist, so these points are not in the domain and, in this book's sense, are not points of discontinuity. The graph does break there: just left of $\frac{\pi}{2}$ ⁠, $\tan x$ is large and positive, and just right of it, large and negative (neither one-sided limit is a real number). But since $\frac{\pi}{2}$ is not in the domain, this does not stop $\tan x$ being a continuous function.

The graph of $y = \tan x$ for $-5.2 \le x \le 5.2$ (horizontal unit 48 px, vertical unit 40 px). Each blue branch is unbroken. The dashed rose lines at $x = \pm\frac{\pi}{2}$ ⁠, $\pm\frac{3\pi}{2}$ mark points outside the domain, where the graph breaks.

Compare Example 12 and $\frac{1}{x}$ in Example 9: a graph in several pieces is fine for a continuous function, as long as the breaks happen only at points where the function is not defined.

Result

$\tan x$ is defined for $x \ne (2n+1)\frac{\pi}{2}$ ⁠, $n \in \Z$ ⁠, and at every such $c$ ⁠, $\lim_{x\to c}\tan x = \tan c$ ⁠. Hence $f(x) = \tan x$ is a continuous function. This agrees with the textbook's solution.

Check it with numbers

At $c = \frac{\pi}{4} \approx 0.785398$ ⁠: $\tan c = 1$ ⁠. With inputs $c \pm 0.001$ ⁠:

\tan(0.784398) \approx 0.998002,

\tan(0.786398) \approx 1.002002,

both within $0.0021$ of 1. Near the excluded point $\frac{\pi}{2} \approx 1.570796$ ⁠: $\tan 1.57 \approx 1255.77$ but $\tan 1.5716 \approx -1244.29$ ⁠, showing the break where $\tan$ is undefined.

Common mistakes

Calling $\frac{\pi}{2}$ a point of discontinuity. It is not in the domain of $\tan x$ ⁠, so it is not a point of discontinuity in this book's sense.
Leaving out the condition $\cos c \ne 0$ ⁠. Theorem 1(4) needs a non-zero denominator at $c$ ⁠; that condition is what limits the result to the domain.
Writing the excluded points as $n\pi$ ⁠. Those are the zeros of $\sin x$ ⁠, where $\tan x = 0$ ⁠. The zeros of $\cos x$ are the odd multiples of $\frac{\pi}{2}$ ⁠.

What to take away

Quotients of continuous functions are continuous wherever the denominator is non-zero: here $\tan = \frac{\sin}{\cos}$ ⁠.
The same reasoning shows $\cot x = \frac{\cos x}{\sin x}$ ⁠, $\sec x = \frac{1}{\cos x}$ and $\csc x = \frac{1}{\sin x}$ are continuous functions (by Theorem 1(4) and Remark (ii)).
Continuous functions can have graphs in several separate pieces, provided the gaps are outside the domain.

Every fact used, and where it comes from

Fact	Used in	Source
$\tan x = \frac{\sin x}{\cos x}$	Steps 1, 3	Class XI, Trigonometric Functions
$\cos x = 0 \iff x = (2n+1)\frac{\pi}{2}$	Steps 1, 5	Class XI, Trigonometric Functions (Basics: trigonometric identities)
Sine and cosine are continuous	Step 2	Example 17 and its Remark, Section 5.2.1
$\frac{f}{g}$ 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)
Continuous function; points outside the domain are not tested	Steps 4, 5	Definitions 1 and 2, Section 5.2 (Lesson: continuity at a point)

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Example 18: tan⁡x\tan xtanx is a continuous function

Basics you need

Solution

Find the domain

Fix a point ccc of the domain

Apply Theorem 1(4)

Conclude with Definition 2

The excluded points

Every fact used, and where it comes from

Example 18: $\tan x$ is a continuous function

Fix a point $c$ of the domain