Question

Evaluate:

$\underset{x\to 0}{lim}\frac{\sin x}{\sqrt{x^2}}=$

• A. $1$
• B. $-1$
• C. $0$
• D. doesn't exist

Answer 1

The correct option is D doesn't exist

Let $L=\underset{x\to 0}{lim}\frac{\sin x}{\sqrt{x^2}}$

To find LHL:

As $x\to 0^{-}\Rightarrow \sin x<0$

But $\sqrt{x^2}>0$

Then, LHL $<0$

As $x\to 0^{+}\Rightarrow \sin x>0$

But $\sqrt{x^2}>0$

Then, RHL $>0$

Thus, LHL $\ne$ RHL.

Hence, the limit does not exist.