The Limit Of Sinx Over X As X Approaches 0

Let x denote the radian measure of an angle. In previous examples, we saw that $lim_{x \to 0} \frac{\sin x}{x} = 1$ . This limit is of importance and we can solve many similar exercises using this limit. In this section, we prove that $lim_{x \to 0} \frac{\sin x}{x} = 1$ using the Sandwich Theorem.

Quick Reference

Limit Formula	Value	Notes / Method
Fundamental Sine Limit	$lim_{x \to 0} \frac{\sin x}{x} = 1$	Angle x in radians
Reciprocal Sine Limit	$lim_{x \to 0} \frac{x}{\sin x} = 1$	Reciprocal of fundamental limit
Scaled Sine Ratio	$lim_{x \to 0} \frac{\sin (A x)}{x} = A$	Substitution $u = A x$
Ratio of Two Sines	$lim_{x \to 0} \frac{\sin (A x)}{\sin (B x)} = \frac{A}{B}$	Divide numerator and denominator by x
Tangent Ratio	$lim_{x \to 0} \frac{\tan x}{x} = 1$	Rewrite $\tan x = \frac{\sin x}{\cos x}$
Cosine Difference Ratios	$lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = \frac{1}{2}$ , $lim_{x \to 0} \frac{1 - \cos x}{x} = 0$	Use half-angle formula $1 - \cos x = 2 \sin^{2} (\frac{x}{2})$
Inverse Sine Ratio	$lim_{x \to 0} \frac{\arcsin x}{x} = 1$	Substitution $u = \arcsin x$

[!NOTE]
Recall that when no $^{\circ}$ notation is present, the default is that x is measured in radians.

The limit of sin(x)/x as x - loading= — The limit of $\frac{\sin x}{x}$ as $x \to 0$ is 1.

Theorem:

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

Proof

Consider the construction shown below, where the radius of the circle is 1 ( $O P = O A = 1$ ), and $0 < x < π / 2$ . It is clear that $\sin x = \frac{P H}{O P} = P H$ and $\tan x = \frac{A T}{O A} = A T$ .

Now notice that:

area (△ O A P) < area (sector O A P) < area (△ O A T)

In triangle OPH, sin x = PH/OP = PH/1 = PH. In triangle OAT, tan x = AT/OA = AT/1 = AT — In $△ O P H$ , $\sin x = \frac{P H}{O P} = P H$ . In $△ O A T$ , $\tan x = \frac{A T}{O A} = A T$ .

Let's find each area in terms of x:

\begin{aligned} \text{area}(\triangle OAP) &= \frac{1}{2}(OA)(PH) = \frac{1}{2}(1)(\sin x) = \frac{1}{2}\sin x \\ \text{area}(\text{sector } OAP) &= \frac{1}{2} r^2 x = \frac{1}{2}(1)^2 x = \frac{x}{2} \\ \text{area}(\triangle OAT) &= \frac{1}{2}(OA)(AT) = \frac{1}{2}(1)(\tan x) = \frac{1}{2}\tan x \end{aligned}

It follows from the above inequalities that:

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

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

Because $0 < x < π / 2$ , $\sin x > 0$ . Dividing each term by $\sin x$ gives:

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

Taking reciprocals reverses the inequalities:

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

Now we show that the last double inequality holds also for $- π / 2 < x < 0$ . If $- π / 2 < x < 0$ , then $0 < - x < π / 2$ and therefore:

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

Recall that $\sin (- x) = - \sin x$ and $\cos (- x) = \cos x$ . Thus:

1 > \frac{- \sin x}{- x} > \cos x ⟹ 1 > \frac{\sin x}{x} > \cos x .

So we have proved:

\cos x < \frac{\sin x}{x} < 1 (- \frac{π}{2} < x < \frac{π}{2}, x \neq 0) .

Because $lim_{x \to 0} \cos x = \cos 0 = 1$ and $lim_{x \to 0} 1 = 1$ , the Sandwich Theorem gives:

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

Applications and Worked Examples

Show that

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

Solution

\begin{aligned} \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. \end{aligned}

Evaluate the following limit:

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

Solution

Let's multiply the 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}

Now let $u = \sqrt{2} x$ . As $x \to 0$ , $u \to 0$ :

\begin{aligned} \lim_{x\to 0} \frac{\sin\sqrt{2}x}{x} &= \lim_{u\to 0} \sqrt{2}\frac{\sin u}{u} \\ &= \sqrt{2} \lim_{u\to 0}\frac{\sin u}{u} \\ &= \sqrt{2} \times 1 = \sqrt{2}. \end{aligned}

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 both numerator and denominator by x:

\begin{aligned} \lim_{x\to 0} \frac{\sin Ax}{\sin Bx} &= \lim_{x\to 0} \frac{\frac{\sin Ax}{x}}{\frac{\sin Bx}{x}} \\ &= \frac{\lim_{x\to 0} \frac{\sin Ax}{x}}{\lim_{x\to 0} \frac{\sin Bx}{x}} \\ &= \frac{A}{B}. \end{aligned}

Show that

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

Solution

By definition $\tan x = \frac{\sin x}{\cos x}$ . Thus:

\begin{aligned} \lim_{x\to 0} \frac{\tan x}{x} &= \lim_{x\to 0} \frac{\frac{\sin x}{\cos x}}{x} \\ &= \lim_{x\to 0} \left( \frac{1}{\cos x} \cdot \frac{\sin x}{x} \right) \\ &= \lim_{x\to 0} \frac{1}{\cos x} \cdot \lim_{x\to 0} \frac{\sin x}{x} \\ &= \frac{1}{\cos 0} \cdot 1 = 1. \end{aligned}

Evaluate the following limit:

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

Solution

Multiply and divide by $\sin x$ :

\begin{aligned} \lim_{x\to 0} \frac{\sin(\sin x)}{x} &= \lim_{x\to 0} \left[ \frac{\sin(\sin x)}{\sin x} \cdot \frac{\sin x}{x} \right] \\ &= \lim_{x\to 0} \frac{\sin(\sin x)}{\sin x} \cdot \lim_{x\to 0} \frac{\sin x}{x} \end{aligned}

For the first limit let $u = \sin x$ . As $x \to 0$ , $u \to 0$ :

\begin{aligned} &= \lim_{u\to 0} \frac{\sin u}{u} \cdot \lim_{x\to 0} \frac{\sin x}{x} \\ &= 1 \cdot 1 = 1. \end{aligned}

Another important limit is $lim_{x \to 0} \frac{1 - \cos x}{x^{2}}$ . Direct substitution of $x = 0$ yields $0 / 0$ . Recall the half-angle formula $1 - \cos x = 2 \sin^{2} (\frac{x}{2})$ :

\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} \\ &= 2 \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) = \frac{1}{2}. \end{aligned}

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

From this limit, we deduce:

lim_{x \to 0} \frac{1 - \cos x}{x} = 0,

since:

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

Find

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

Solution

Using $\tan x = \frac{\sin x}{\cos x}$ :

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

Show that

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

Solution

We rewrite the expression as:

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

Find

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

Solution

Let $u = \arcsin x ⟹ \sin u = x$ . As $x \to 0$ , $u \to 0$ :

\begin{aligned} \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{\lim_{u\to 0} 1}{\lim_{u\to 0} \frac{\sin u}{u}} \\ &= \frac{1}{1} = 1. \end{aligned}

Exercises

Evaluate the limit or show that it does not exist:

lim_{x \to 0} \frac{x}{\sin (3 x)}

Answer

$\frac{1}{3}$

Solution

Multiply and divide by $3$ :

\begin{aligned} \lim_{x\to 0} \frac{x}{\sin(3x)} &= \lim_{x\to 0} \left[ \frac{1}{3} \cdot \frac{3x}{\sin(3x)} \right] \\ &= \frac{1}{3} \lim_{x\to 0} \frac{1}{\frac{\sin(3x)}{3x}} \\ &= \frac{1}{3} \cdot \frac{1}{1} = \frac{1}{3}. \end{aligned}

Evaluate the limit or show that it does not exist:

lim_{x \to 0} \frac{4 x}{\sin (7 x)}

Answer

$\frac{4}{7}$

Solution

Rewrite the fraction:

\begin{aligned} \lim_{x\to 0} \frac{4x}{\sin(7x)} &= \lim_{x\to 0} \left[ \frac{4}{7} \cdot \frac{7x}{\sin(7x)} \right] \\ &= \frac{4}{7} \lim_{x\to 0} \frac{1}{\frac{\sin(7x)}{7x}} \\ &= \frac{4}{7} \cdot 1 = \frac{4}{7}. \end{aligned}

Evaluate the limit or show that it does not exist:

lim_{t \to 0} \frac{\tan (4 t)}{16 t}

Answer

$\frac{1}{4}$

Solution

Use $\tan (4 t) = \frac{\sin (4 t)}{\cos (4 t)}$ :

\begin{aligned} \lim_{t\to 0} \frac{\tan(4t)}{16t} &= \lim_{t\to 0} \left[ \frac{\sin(4t)}{16t} \cdot \frac{1}{\cos(4t)} \right] \\ &= \lim_{t\to 0} \left[ \frac{1}{4} \cdot \frac{\sin(4t)}{4t} \cdot \frac{1}{\cos(4t)} \right] \\ &= \frac{1}{4} \cdot (1) \cdot \frac{1}{\cos 0} = \frac{1}{4}. \end{aligned}

Evaluate the limit or show that it does not exist:

lim_{θ \to 0} \frac{\sec (2 θ) \tan (5 θ)}{θ}

Answer

$5$

Solution

Express $\sec (2 θ) = \frac{1}{\cos (2 θ)}$ and $\tan (5 θ) = \frac{\sin (5 θ)}{\cos (5 θ)}$ :

\begin{aligned} \lim_{\theta\to 0} \frac{\sec(2\theta)\tan(5\theta)}{\theta} &= \lim_{\theta\to 0} \left[ \sec(2\theta) \cdot \frac{1}{\cos(5\theta)} \cdot \frac{\sin(5\theta)}{\theta} \right] \\ &= \lim_{\theta\to 0} \left[ \sec(2\theta) \cdot \frac{1}{\cos(5\theta)} \cdot 5 \cdot \frac{\sin(5\theta)}{5\theta} \right] \\ &= (1) \cdot (1) \cdot 5 \cdot (1) = 5. \end{aligned}

Evaluate the limit or show that it does not exist:

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

Answer

$\frac{3}{5}$

Solution

Divide numerator and denominator by $x$ :

\begin{aligned} \lim_{x\to 0} \frac{\tan(3x)}{\sin(5x)} &= \lim_{x\to 0} \left[ \frac{\sin(3x)}{\sin(5x)} \cdot \frac{1}{\cos(3x)} \right] \\ &= \lim_{x\to 0} \left[ \frac{\frac{\sin(3x)}{x}}{\frac{\sin(5x)}{x}} \cdot \frac{1}{\cos(3x)} \right] \\ &= \frac{\lim_{x\to 0} \left(3 \cdot \frac{\sin(3x)}{3x}\right)}{\lim_{x\to 0} \left(5 \cdot \frac{\sin(5x)}{5x}\right)} \cdot \lim_{x\to 0} \frac{1}{\cos(3x)} \\ &= \frac{3(1)}{5(1)} \cdot 1 = \frac{3}{5}. \end{aligned}

Evaluate the limit or show that it does not exist:

lim_{t \to π / 2} \frac{\cos t}{\frac{π}{2} - t}

(Hint: let $θ = \frac{π}{2} - t$ )

Answer

$1$

Solution

Let $θ = \frac{π}{2} - t$ . As $t \to \frac{π}{2}$ , we have $θ \to 0$ , and $t = \frac{π}{2} - θ$ .

Using the co-function identity $\cos t = \cos (\frac{π}{2} - θ) = \sin θ$ :

lim_{t \to π / 2} \frac{\cos t}{\frac{π}{2} - t} = lim_{θ \to 0} \frac{\sin θ}{θ} = 1.

Evaluate the limit or show that it does not exist:

lim_{x \to 0} \frac{x \sin x}{1 - \cos x}

Answer

$2$

Solution

Multiply numerator and denominator by the conjugate $(1 + \cos x)$ :

\begin{aligned} \lim_{x\to 0} \frac{x\sin x}{1 - \cos x} &= \lim_{x\to 0} \frac{x\sin x (1 + \cos x)}{(1 - \cos x)(1 + \cos x)} \\ &= \lim_{x\to 0} \frac{x\sin x (1 + \cos x)}{1 - \cos^2 x} \\ &= \lim_{x\to 0} \frac{x\sin x (1 + \cos x)}{\sin^2 x} \\ &= \lim_{x\to 0} \left[ \frac{x}{\sin x} \cdot (1 + \cos x) \right] \\ &= \left( \lim_{x\to 0} \frac{x}{\sin x} \right) \left( \lim_{x\to 0} (1 + \cos x) \right) \\ &= (1) \cdot (1 + 1) = 2. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{1 - \cos x}{1 - \cos (\frac{x}{2})}

(Hint: Remember the half-angle identity $\sin^{2} θ = \frac{1}{2} [1 - \cos (2 θ)]$ .)

Answer

$4$

Solution

Using the half-angle identity $1 - \cos (2 θ) = 2 \sin^{2} θ$ , we can express the numerator and denominator as:

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

1 - \cos (\frac{x}{2}) = 2 \sin^{2} (\frac{x}{4}) .

Substituting these expressions into the limit:

\begin{aligned} \lim_{x\to 0} \frac{1 - \cos x}{1 - \cos\left(\frac{x}{2}\right)} &= \lim_{x\to 0} \frac{2\sin^2\left(\frac{x}{2}\right)}{2\sin^2\left(\frac{x}{4}\right)} \\ &= \lim_{x\to 0} \left[ \frac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2} \cdot \frac{\left(\frac{x}{4}\right)^2}{\sin^2\left(\frac{x}{4}\right)} \cdot \frac{\left(\frac{x}{2}\right)^2}{\left(\frac{x}{4}\right)^2} \right] \\ &= (1)^2 \cdot (1)^2 \cdot \frac{\frac{x^2}{4}}{\frac{x^2}{16}} \\ &= \frac{16}{4} = 4. \end{aligned}
Alternative Method: Using the identity

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

lim_{x \to 0} \frac{2 (1 - \cos (\frac{x}{2})) (1 + \cos (\frac{x}{2}))}{1 - \cos (\frac{x}{2})} = lim_{x \to 0} 2 (1 + \cos (\frac{x}{2})) = 2 (1 + 1) = 4.

Evaluate the limit:

lim_{x \to 0} \frac{\arctan x}{x} (where \arctan x = \tan^{- 1} x)

Answer

$1$

Solution

Let $θ = \arctan x$ . Then $x = \tan θ$ . As $x \to 0$ , we have $θ = \arctan (0) \to 0$ .

Substituting $θ$ into the limit:

\begin{aligned} \lim_{x\to 0} \frac{\arctan x}{x} &= \lim_{\theta\to 0} \frac{\theta}{\tan\theta} \\ &= \lim_{\theta\to 0} \frac{\theta}{\frac{\sin\theta}{\cos\theta}} \\ &= \lim_{\theta\to 0} \left( \frac{\theta}{\sin\theta} \cdot \cos\theta \right) \\ &= (1) \cdot (1) = 1. \end{aligned}

Evaluate the limit:

lim_{x \to 0} \frac{\sin (3 x) + \arctan^{2} x - \arcsin^{2} x}{4 x}

Answer

$\frac{3}{4}$

Solution

Split the fraction into three separate terms:

\frac{\sin (3 x) + \arctan^{2} x - \arcsin^{2} x}{4 x} = \frac{\sin (3 x)}{4 x} + \frac{\arctan^{2} x}{4 x} - \frac{\arcsin^{2} x}{4 x} .

Taking the limit of each term as $x \to 0$ :

First term: $lim_{x \to 0} \frac{\sin (3 x)}{4 x} = \frac{3}{4} lim_{x \to 0} \frac{\sin (3 x)}{3 x} = \frac{3}{4} (1) = \frac{3}{4} .$
Second term: $lim_{x \to 0} \frac{\arctan^{2} x}{4 x} = \frac{1}{4} lim_{x \to 0} (\frac{\arctan x}{x} \cdot \arctan x) = \frac{1}{4} (1 \cdot 0) = 0.$
Third term: $lim_{x \to 0} \frac{\arcsin^{2} x}{4 x} = \frac{1}{4} lim_{x \to 0} (\frac{\arcsin x}{x} \cdot \arcsin x) = \frac{1}{4} (1 \cdot 0) = 0.$ Combining the three limits: $lim_{x \to 0} \frac{\sin (3 x) + \arctan^{2} x - \arcsin^{2} x}{4 x} = \frac{3}{4} + 0 - 0 = \frac{3}{4} .$

Evaluate the limit:

lim_{x \to 0} \frac{\tan (2 x) - x + x^{2}}{\sin x + 2 \sin^{2} (4 x) + 5 x^{3}}

Answer

$1$

Solution

Divide both the numerator and denominator by $x$ (for $x \neq 0$ ):

lim_{x \to 0} \frac{\frac{\tan (2 x) - x + x^{2}}{x}}{\frac{\sin x + 2 \sin^{2} (4 x) + 5 x^{3}}{x}} = lim_{x \to 0} \frac{\frac{\tan (2 x)}{x} - 1 + x}{\frac{\sin x}{x} + 2 \cdot \frac{\sin^{2} (4 x)}{x} + 5 x^{2}} .

Evaluate the limit of the numerator as $x \to 0$ :

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

Evaluate the limit of the denominator as $x \to 0$ :

lim_{x \to 0} (\frac{\sin x}{x} + 32 x \cdot \frac{\sin^{2} (4 x)}{(4 x)^{2}} + 5 x^{2}) = 1 + 0 (1) + 0 = 1.

Therefore:

lim_{x \to 0} \frac{\tan (2 x) - x + x^{2}}{\sin x + 2 \sin^{2} (4 x) + 5 x^{3}} = \frac{1}{1} = 1.

Evaluate the limit:

lim_{x \to 0^{+}} \frac{\arccos (1 - x)}{\sqrt{x}}

(Hint: Let $θ = \arccos (1 - x)$ and use the identity $1 - \cos θ = 2 \sin^{2} (\frac{θ}{2})$ .)

Answer

$\sqrt{2}$

Solution

Let $θ = \arccos (1 - x)$ . Then $\cos θ = 1 - x ⟹ x = 1 - \cos θ$ .

As $x \to 0^{+}$ , we have $\cos θ \to 1^{-}$ , so $θ \to 0^{+}$ .

Using the half-angle identity $1 - \cos θ = 2 \sin^{2} (\frac{θ}{2})$ :

\sqrt{x} = \sqrt{1 - \cos θ} = \sqrt{2 \sin^{2} (\frac{θ}{2})} = \sqrt{2} \sin (\frac{θ}{2}) (since \frac{θ}{2} > 0) .

Substituting $θ$ and $\sqrt{x}$ into the limit:

\begin{aligned} \lim_{x\to 0^+} \frac{\arccos(1-x)}{\sqrt{x}} &= \lim_{\theta\to 0^+} \frac{\theta}{\sqrt{2}\sin\left(\frac{\theta}{2}\right)} \\ &= \frac{2}{\sqrt{2}} \lim_{\theta\to 0^+} \frac{\frac{\theta}{2}}{\sin\left(\frac{\theta}{2}\right)} \\ &= \sqrt{2} \cdot 1 = \sqrt{2}. \end{aligned}

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