Trigonometric and Inverse Trigonometric Functions

Lessons

Senior 5 · Mathematics

Trigonometric and Inverse Trigonometric Functions

5 lessons

Lesson 2 of 5

Limits of trigonometric and inverse trigonometric functions

25 min

A limit the calculator only hints at

Put $x = 0.5$ , $0.1$ and $0.01$ (in radians) into $\frac{\sin x}{x}$ : you get $0.958\,851$ , $0.998\,334$ and $0.999\,983$ . The values creep up to 1, yet at $x = 0$ the fraction is $\frac{0}{0}$ and means nothing. In S4 you met limits of polynomial and rational functions, and the squeeze (sandwich) theorem. This lesson proves the one new limit that the whole of trigonometric calculus rests on, and derives the others from it.

Important

The fundamental trigonometric limit ( $x$ in radians)

\lim_{x \to 0}\frac{\sin x}{x} = 1.

Proof. Take $0 \lt x \lt \frac{\pi}{2}$ (Figure 1). Comparing areas, $\frac{1}{2}\sin x \lt \frac{1}{2}x \lt \frac{1}{2}\tan x$ . Divide by $\frac{1}{2}\sin x \gt 0$ : $1 \lt \frac{x}{\sin x} \lt \frac{1}{\cos x}$ , and take reciprocals:

\cos x \lt \frac{\sin x}{x} \lt 1.

As $x \to 0^+$ , $\cos x \to 1$ , so by the squeeze theorem $\frac{\sin x}{x} \to 1$ . Since $\frac{\sin x}{x}$ is even, the same holds as $x \to 0^-$ . $\blacksquare$

Figure 1. For

0 \lt x \lt \frac{\pi}{2}

: triangle

O A P

\subset

sector

O A P

\subset

triangle

O A T

, so

\frac{1}{2}\sin x \lt \frac{1}{2}x \lt \frac{1}{2}\tan x

Limits that follow

$\lim_{x \to 0}\frac{\tan x}{x} = \lim_{x \to 0}\frac{\sin x}{x}\cdot\frac{1}{\cos x} = 1$ .
$\lim_{x \to 0}\frac{\sin kx}{x} = k$ : write it as $k\cdot\frac{\sin kx}{kx}$ and let $u = kx \to 0$ .
$\lim_{x \to 0}\frac{1 - \cos x}{x^2} = \frac{1}{2}$ : $\frac{1 - \cos x}{x^2} = \frac{2\sin^2\frac{x}{2}}{x^2} = \frac{1}{2}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \to \frac{1}{2}$ . Hence also $\frac{1 - \cos x}{x} = x\cdot\frac{1 - \cos x}{x^2} \to 0$ .
$\lim_{x \to 0}\frac{\arcsin x}{x} = 1$ and $\lim_{x \to 0}\frac{\arctan x}{x} = 1$ : put $t = \arcsin x$ , so $x = \sin t$ and $\frac{t}{\sin t} \to 1$ as $t \to 0$ ; the same with $\tan$ .

At infinity. $\sin x$ and $\cos x$ have no limit as $x \to \infty$ (they keep oscillating), but they are bounded, which is often enough: $-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$ for $x \gt 0$ gives $\frac{\sin x}{x} \to 0$ . And from Figure 2 of Lesson 1, $\lim_{x \to +\infty}\arctan x = \frac{\pi}{2}$ and $\lim_{x \to -\infty}\arctan x = -\frac{\pi}{2}$ .

Continuity. The six trigonometric functions and the inverse functions are continuous at every point of their domains, so for example $\lim_{x \to \frac{\pi}{3}}\cos x = \frac{1}{2}$ . At the edges of the domain of $\tan$ : $\lim_{x \to \frac{\pi}{2}^-}\tan x = +\infty$ and $\lim_{x \to \frac{\pi}{2}^+}\tan x = -\infty$ .

Figure 2.

y = \frac{\sin x}{x}

is even, undefined at 0, and tends to 1 as

x \to 0

. For large

∣ x ∣

it is squeezed between

\pm\frac{1}{|x|}

, so it tends to 0.

Definition

Squeeze (sandwich) theorem: if $g(x) \le f(x) \le h(x)$ near $a$ (except perhaps at $a$ ), and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$ , then $\lim_{x \to a} f(x) = L$ . The same holds as $x \to \pm\infty$ .

Example: $-x^2 \le x^2\sin\frac{1}{x} \le x^2$ , and both bounds tend to 0, so $\lim_{x \to 0} x^2\sin\frac{1}{x} = 0$ .

Worked example 1: using the standard limits

Find (a) $\lim_{x \to 0}\frac{\sin 5x}{3x}$ (b) $\lim_{x \to 0}\frac{1 - \cos 4x}{x^2}$ (c) $\lim_{x \to 0}\frac{\tan 3x}{\sin 2x}$ (d) $\lim_{x \to 0}\frac{\arcsin 3x}{x}$ .

Solution

$\frac{\sin 5x}{3x} = \frac{5}{3}\cdot\frac{\sin 5x}{5x} \to \frac{5}{3}$ .
$1 - \cos 4x = 2\sin^2 2x$ , so $\frac{1 - \cos 4x}{x^2} = 2\cdot 4\cdot\left(\frac{\sin 2x}{2x}\right)^2 \to 8$ .
$\frac{\tan 3x}{\sin 2x} = \frac{\tan 3x}{3x}\cdot\frac{2x}{\sin 2x}\cdot\frac{3}{2} \to 1 \cdot 1 \cdot \frac{3}{2} = \frac{3}{2}$ .
$\frac{\arcsin 3x}{x} = 3\cdot\frac{\arcsin 3x}{3x} \to 3$ .

Worked example 2: shift first, or squeeze

Find (a) $\lim_{x \to \pi}\frac{\sin x}{x - \pi}$ (b) $\lim_{x \to +\infty}\frac{2x + \cos x}{x}$ (c) $\lim_{x \to +\infty} x\sin\frac{1}{x}$ .

Solution

Put $t = x - \pi \to 0$ : $\sin(t + \pi) = -\sin t$ , so the limit is $\lim_{t \to 0}\frac{-\sin t}{t} = -1$ .
$\frac{2x + \cos x}{x} = 2 + \frac{\cos x}{x}$ , and $-\frac{1}{x} \le \frac{\cos x}{x} \le \frac{1}{x} \to 0$ . The limit is 2.
Put $t = \frac{1}{x} \to 0^+$ : $x\sin\frac{1}{x} = \frac{\sin t}{t} \to 1$ .

Warning

Radians only. $\frac{\sin x}{x} \to 1$ is true only when $x$ is in radians: in degrees, $\frac{\sin x^\circ}{x} \to \frac{\pi}{180}$ . And the limit is at $x \to 0$ : $\frac{\sin x}{x}$ does not tend to 1 as $x \to \infty$ (it tends to 0), and $\frac{\sin 2x}{x}$ tends to 2, not 1.

Tip

Make the pattern appear. Every limit here is solved by arranging $\frac{\sin(\text{something})}{\text{the same something}}$ , with the "something" tending to 0. Multiply and divide by the constant you need, as in $\frac{\sin 5x}{3x} = \frac{5}{3}\cdot\frac{\sin 5x}{5x}$ .

Khan Academy: Proof: lim (sin x)/x. Watch how the areas of two triangles and a sector of the unit circle squeeze sin x over x between cos x and 1.

✦

Practice

The last two are harder.

$\lim_{x \to 0}\frac{\sin 7x}{x}$
$\lim_{x \to 0}\frac{\sin 3x}{\sin 5x}$
$\lim_{x \to 0}\frac{1 - \cos x}{x\sin x}$
$\lim_{x \to 0}\frac{\tan x - \sin x}{x^3}$
$\lim_{x \to \frac{\pi}{2}}\frac{\cos x}{x - \frac{\pi}{2}}$
$\lim_{x \to +\infty}\frac{3x - \sin x}{x + 2}$
$\lim_{x \to 0}\frac{\arctan 2x}{\sin 3x}$
(a) Show that $\lim_{x \to 0} x\sin\frac{1}{x} = 0$ . (b) Explain why $\lim_{x \to 0}\arctan\frac{1}{x}$ does not exist.

Show answersHide answers

7.
$\frac{\sin 3x}{3x}\cdot\frac{5x}{\sin 5x}\cdot\frac{3}{5} \to \frac{3}{5}$ .
$\frac{1 - \cos x}{x^2}\cdot\frac{x}{\sin x} \to \frac{1}{2}\cdot 1 = \frac{1}{2}$ .
$\tan x - \sin x = \tan x\,(1 - \cos x)$ , so the fraction is $\frac{\tan x}{x}\cdot\frac{1 - \cos x}{x^2} \to \frac{1}{2}$ .
Put $t = x - \frac{\pi}{2}$ : $\cos\left(t + \frac{\pi}{2}\right) = -\sin t$ , so the limit is $- 1$ .
Divide by $x$ : $\frac{3 - \frac{\sin x}{x}}{1 + \frac{2}{x}} \to \frac{3 - 0}{1 + 0} = 3$ .
$\frac{\arctan 2x}{2x}\cdot\frac{3x}{\sin 3x}\cdot\frac{2}{3} \to \frac{2}{3}$ .
(a) $-|x| \le x\sin\frac{1}{x} \le |x|$ and $|x| \to 0$ . (b) As $x \to 0^+$ , $\frac{1}{x} \to +\infty$ , so $\arctan\frac{1}{x} \to \frac{\pi}{2}$ ; as $x \to 0^-$ , it tends to $-\frac{\pi}{2}$ . The one-sided limits differ.

Sources and credits

Written by Ganzaa: the explanations, the proof, the diagrams, the worked examples, the practice items and the quiz questions.
Video: Proof: lim (sin x)/x, Khan Academy, embedded from YouTube.

1A limit the calculator only hints at

2Limits that follow

A limit the calculator only hints at

Limits that follow