Question

$\underset{x\to 0}{\lim }\frac{\left|\sin x\right|}{x}$ is

(a) 1 (b) −1 (c) 0 (d) does not exist

Answer 1

We have,

$\left|\sin x\right|=\left\{\begin{array}{ll}\sin x,& 0\le x\le \frac{\mathrm{\pi }}{2}\\ -\sin x,& -\frac{\mathrm{\pi }}{2}\le x<0\end{array}\right.$

Now,

$\underset{x\to 0^{-}}{\lim }\frac{\left|\sin x\right|}{x}=\underset{x\to 0}{\lim }\frac{-\sin x}{x}=-\underset{x\to 0}{\lim }\frac{\sin x}{x}=-1$

$\underset{x\to 0^{+}}{\lim }\frac{\left|\sin x\right|}{x}=\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$

Clearly, $\underset{x\to 0^{-}}{\lim }\frac{\left|\sin x\right|}{x}\ne \underset{x\to 0^{+}}{\lim }\frac{\left|\sin x\right|}{x}$

∴ $\underset{x\to 0}{\lim }\frac{\left|\sin x\right|}{x}$ does not exist.

Hence, the correct answer is option (d).


Note: Here, a small interval close to 0 is only considered.