The squeeze theorem and the limit of sin x / x

Warm-up

Three quick questions from earlier lessons.

Discover

With $x$ in radians: $\dfrac{\sin 0.1}{0.1} \approx 0.9983$ , $\dfrac{\sin 0.01}{0.01} \approx 0.99998$ . What does $\dfrac{\sin x}{x}$ approach as $x \to 0$ ?

1
0

Key idea

Two tools for limits that algebra cannot simplify.

\displaystyle\lim_{u \to 0} \dfrac{\sin u}{u} = 1

(

u

in radians), so

\dfrac{\sin(ax)}{x} \to a

. Squeeze theorem: if

g(x) \le f(x) \le h(x)

near

a

and

g

and

h

have the same limit

L

at

a

, then

\displaystyle\lim_{x \to a} f(x) = L

.

Worked example

Find $\displaystyle\lim_{x \to 0} \dfrac{\sin(5x)}{x}$ .

\dfrac{\sin(5x)}{x} = 5 \cdot \dfrac{\sin(5x)}{5x}

Make the denominator match the angle 5x.

\to 5 \cdot 1

u = 5x → 0, and sin(u)/u → 1.

= 5

The limit.

Your turn, guided

Your turn: $\displaystyle\lim_{x \to 0} \dfrac{\sin(3x)}{2x}$ .

= \dfrac{3}{2} \cdot \dfrac{\sin(3x)}{3x}

Write it with sin(3x)/(3x).

\to \dfrac{3}{2}

The ratio tends to 1.

Practice ★

Find $\displaystyle\lim_{x \to 0} \dfrac{\sin(6x)}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{\sin(2x)}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{\sin(9x)}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{\sin(7x)}{x}$ .

Watch out!

Practice ★★

Find $\displaystyle\lim_{x \to 0} \dfrac{\sin(3x)}{\sin(9x)}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{\sin(8x)}{\sin(9x)}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{\sin(4x)}{\sin(8x)}$ .

What is $\displaystyle\lim_{x \to 0} x^{ 2 }\sin\left(\dfrac{1}{x}\right)$ ?
1. 0, by the squeeze theorem
2. It does not exist, because $\sin(1/x)$ oscillates
3. 1

Exit ticket

Three questions to check you have got it.

Find $\displaystyle\lim_{x \to 0} \dfrac{\sin(6x)}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{\sin(2x)}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{\sin(9x)}{x}$ .

Summary card

Special limits

$\dfrac{\sin u}{u} \to 1$ as $u \to 0$ (radians); so $\dfrac{\sin(ax)}{x} \to a$ .
Squeeze: trapped between two functions with the same limit.

Watch out: $\dfrac{\sin(ax)}{x}$ tends to $a$ , not 1.

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