Answer to Question #201494 in Calculus for Sanjeshni

Question

Use the Squeezing Theorem to show that

lim

X approaches to +sum to infinity

Sin x divide by x =0

Answer 1

$$-1\leq\sin x\leq 1$$

If $x>0$

$$-\dfrac{1}{x}\leq\dfrac{\sin x}{x}\leq \dfrac{1}{x}$$

$$\lim\limits_{x\to+ \infin}(-\dfrac{1}{x})=0\ and\ \lim\limits_{x\to +\infin}(\dfrac{1}{x})=0$$

Therefore by the Squeezing Theorem

$$\lim\limits_{x\to + \infin}(\dfrac{\sin x}{x})=0$$