limit of ${\displaystyle \frac{sinx}{x}}$ as $x$ approaches 0

Theorem 1.

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=1$$

for .

Proof.

First, let . Then . Note also that

(1)

Multiplying both of this inequality by $\cos x$ yields

(2)

By this theorem (http://planetmath.org/ComparisonOfSinThetaAndThetaNearTheta0),

(3)

Combining inequalities (2) and (3) gives

(4)

Dividing by $x$ yields

(5)

Now let . Then . Plugging $-x$ into inequality (5) gives

(6)

Since $\cos$ is an even function and $\sin$ is an odd function, we have

(7)

Therefore, inequality (5) holds for all real $x$ with .

Since $\cos$ is continuous, $\underset{x\to 0}{lim}\cos x=\cos 0=1$. Thus,

$$1=\underset{x\to 0}{lim}\cos x\le \underset{x\to 0}{lim}\frac{\sin x}{x}\le \underset{x\to 0}{lim}1=1.$$(8)

By the squeeze theorem, it follows that $\underset{x\to 0}{lim}{\displaystyle \frac{\sin x}{x}}=1$.∎

Note that the above limit is also valid if $x$ is considered as a complex variable.