Mathematics - Limits Question with Solution | TestHub

Let $l={\lim }_{x\to 0}\frac{[\sin x]}{x};\mathrm{RHL}={\lim }_{x\to 0^{+}}\frac{\sin x}{x}=1$
$\mathrm{LHL}={\lim }_{x\to 0^{-}}\left(\frac{-\sin x}{x}\right)=-\left({\lim }_{x\to 0^{-}}\frac{\sin x}{x}\right)=-1$
$\left[|x|=\begin{array}{c}x\text{ if }x\ge 0\\ -x\text{ if }x\le 0\end{array}\right]$
As, LHL $\ne$ RHL Hence, limit $l$.does not exist.