$f(x)=\left\{\begin{array}{l}\frac{\sin [x]}{[x]}, &[x] \neq 0 \\ 0,& [x]=0\end{array}\right.$, then $\lim\limits_{x \rightarrow 0} f(x)$ is
Answer & explanation
Correct answer: option 4
$f(x)=\left\{\begin{array}{l}\frac{\sin [x]}{[x]} & [x] \neq 0 \\ 0 & [x]=0\end{array}\right.$
$\lim\limits_{x \rightarrow 0^{-}} \frac{\sin [x]}{[x]}=\frac{\sin -1}{-1}=\sin 1=LHL$
$\lim\limits_{x \rightarrow 0^{+}} \frac{\sin [x]}{[x]}=\frac{\sin 0}{0}$ (limit doesn't exist)