The Limit of sin(x)/x as x→0

The limit $lim_{x \to 0} \frac{\sin x}{x} = 1$ is one of the most important results in calculus. It is the key to differentiating trigonometric functions, and it underlies many related limits. This section gives a geometric proof and develops a family of related results.

Limit	Value
$lim_{x \to 0} \frac{\sin x}{x}$	$1$
$lim_{x \to 0} \frac{x}{\sin x}$	$1$
$lim_{x \to 0} \frac{\sin A x}{x}$	$A$
$lim_{x \to 0} \frac{\sin A x}{\sin B x}$	$A / B$
$lim_{x \to 0} \frac{\tan x}{x}$	$1$
$lim_{x \to 0} \frac{1 - \cos x}{x^{2}}$	$1 / 2$
$lim_{x \to 0} \frac{1 - \cos x}{x}$	$0$

Recall: when no degree symbol $^{\circ}$ appears, angles are measured in radians.

The Main Theorem

Graph of y=sin(x)/x showing the function approaches 1 as x approaches 0

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

Proof

Consider the construction below, where the circle has radius $1$ (

O P = O A = 1

) and

0 < x < π / 2

. From the geometry:

\sin x = P H, \tan x = A T .

Geometric diagram with unit circle, sector OAP, and triangles used to prove the sinc limit

Comparing areas:

area of △ O A P < area of sector O A P < area of △ O A T .

Computing each: \begin{aligned} \text{area of } \triangle OAP &= \tfrac{1}{2}(1)(\sin x) = \tfrac{1}{2}\sin x \\ \text{area of sector } OAP &= \tfrac{1}{2}r^2\theta = \tfrac{1}{2}(1)x = \tfrac{x}{2} \\ \text{area of } \triangle OAT &= \tfrac{1}{2}(1)(\tan x) = \tfrac{1}{2}\tan x \end{aligned} So

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

, or equivalently

\sin x < x < \frac{\sin x}{\cos x}

. Since

0 < x < π / 2

\sin x > 0

. Dividing by

\sin x

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

Taking reciprocals (reversing inequalities): \cos x < \frac{\sin x}{x} < 1. \tag{i} Now show (i) holds also for

- π / 2 < x < 0

. If

- π / 2 < x < 0

then

0 < - x < π / 2

, so from (i):

1 > \frac{\sin (- x)}{- x} > \cos (- x) .

Using

\sin (- x) = - \sin x

and

\cos (- x) = \cos x

, this gives

\cos x < \frac{\sin x}{x} < 1

as well. Thus for all

x \in (- π / 2, π / 2)

x \neq 0

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

Since

lim_{x \to 0} \cos x = 1

and

lim_{x \to 0} 1 = 1

, the Sandwich Theorem gives:

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

Related Limits and Examples

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

Solution

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

Evaluate $lim_{x \to 0} \frac{\sin (\sqrt{2} x)}{x}$ .

Solution

Multiply numerator and denominator by

\sqrt{2}

lim_{x \to 0} \frac{\sin \sqrt{2} x}{x} = lim_{x \to 0} \frac{\sqrt{2} \sin (\sqrt{2} x)}{\sqrt{2} x} .

Let

u = \sqrt{2} x

; as

x \to 0

u \to 0

= lim_{u \to 0} \sqrt{2} \frac{\sin u}{u} = \sqrt{2} \cdot 1 = \sqrt{2} .

In general, for any constant $A$ :

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

Show that for nonzero constants $A$ and $B$ : $lim_{x \to 0} \frac{\sin A x}{\sin B x} = \frac{A}{B}$ .

Solution

Divide numerator and denominator by

x

lim_{x \to 0} \frac{\sin A x}{\sin B x} = lim_{x \to 0} \frac{\frac{\sin A x}{x}}{\frac{\sin B x}{x}} = \frac{lim_{x \to 0} \frac{\sin A x}{x}}{lim_{x \to 0} \frac{\sin B x}{x}} = \frac{A}{B} .

Show that $lim_{x \to 0} \frac{\tan x}{x} = 1$ .

Solution

Since

\tan x = \sin x / \cos x

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

Evaluate $lim_{x \to 0} \frac{\sin (\sin x)}{x}$ .

Solution

Multiply and divide by

\sin x

lim_{x \to 0} \frac{\sin (\sin x)}{x} = lim_{x \to 0} \frac{\sin (\sin x)}{\sin x} \cdot \frac{\sin x}{x} .

Let

u = \sin x

; as

x \to 0

u = \sin x \to 0

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

The Limit of $(1 - \cos x) / x^{2}$

Notice that direct substitution of $x = 0$ gives $0 / 0$ . Using the half-angle formula $\frac{1 - \cos 2 θ}{2} = \sin^{2} θ$ with $x = 2 θ$ :

1 - \cos x = 2 \sin^{2} (\frac{x}{2}) .

Therefore:

\begin{aligned} \lim_{x \to 0} \frac{1-\cos x}{x^2} &= \lim_{x \to 0} \frac{2\sin^2(x/2)}{x^2} = 2\lim_{x \to 0} \frac{\sin(x/2)}{x} \cdot \lim_{x \to 0} \frac{\sin(x/2)}{x}. \end{aligned}

Since $lim_{x \to 0} \frac{\sin (A x)}{x} = A$ , with $A = 1 / 2$ :

lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = 2 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2} .

It also follows that:

lim_{x \to 0} \frac{1 - \cos x}{x} = lim_{x \to 0} (x \cdot \frac{1 - \cos x}{x^{2}}) = 0 \cdot \frac{1}{2} = 0.

Find $lim_{x \to 0} \frac{\tan x - \sin x}{x^{3}}$ .

Solution

Using

\tan x = \sin x / \cos x

: \begin{aligned} \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} &= \lim_{x \to 0} \frac{\dfrac{\sin x}{\cos x} - \sin x}{x^3} = \lim_{x \to 0} \frac{\sin x(1-\cos x)}{\cos x \cdot x^3} \\ &= \lim_{x \to 0} \frac{1}{\cos x} \cdot \frac{\sin x}{x} \cdot \frac{1-\cos x}{x^2} \\ &= \frac{1}{\cos 0} \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}. \end{aligned}

Show that $lim_{x \to 0} \frac{x^{2} \sin (1 / x)}{\sin x} = 0$ .

Solution

Rewrite the expression:

\frac{x^{2} \sin (1 / x)}{\sin x} = \frac{x}{\sin x} \cdot x \sin \frac{1}{x} .

Taking limits:

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

Find $lim_{x \to 0} \frac{\arcsin x}{x}$ .

Solution

Let

u = \arcsin x

, so

\sin u = x

. As

x \to 0

u = \arcsin x \to 0

. Therefore:

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

Frequently Asked Questions

Why must

x

be in radians for

lim_{x \to 0} \sin x / x = 1

The proof uses the area of a circular sector, which equals

\frac{1}{2} r^{2} θ

only when

θ

is in radians. In degrees, the arc length formula changes, and the limit becomes

π / 180

instead of $1$.

How do I handle

lim_{x \to 0} \sin (3 x) / \sin (5 x)

Use

lim_{x \to 0} \sin (A x) / \sin (B x) = A / B

. So

lim_{x \to 0} \sin (3 x) / \sin (5 x) = 3 / 5

. Alternatively, divide numerator and denominator by

x

and use the fact that

lim_{x \to 0} \sin (A x) / x = A

for each piece.

What is

lim_{x \to 0} (1 - \cos x) / x

? Why does it differ from

(1 - \cos x) / x^{2}

lim_{x \to 0} (1 - \cos x) / x = 0

, while

lim_{x \to 0} (1 - \cos x) / x^{2} = 1 / 2

. The numerator

1 - \cos x

behaves like

x^{2} / 2

near

x = 0

, so dividing by

x

gives something approaching zero, while dividing by

x^{2}

gives something approaching

1 / 2

The Main Theorem

Related Limits and Examples

The Limit of $(1 - \cos x) / x^{2}$

Frequently Asked Questions

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The Limit of sin(x)/x as x→0

The Main Theorem

Related Limits and Examples

The Limit of ( 1 − cos ⁡ x ) / x 2

Frequently Asked Questions

Quick Links

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The Limit of $(1 - \cos x) / x^{2}$