If  $f\left(x\right)=\left\{\begin{array}{cc}\frac{\sin \left[x\right]}{\left[x\right]},& \text{ for }\left[x\right]\ne 0\\ 0,& \text{ for }\left[x\right]=0\end{array}\right.$ where $\left[x\right]$ denotes the greatest integer less than or equal to $x$, then $\underset{x\to 0}{lim} f\left(x\right)$

 

The given funciton is 

$f\left(x\right)=\left\{\begin{array}{ccc}\frac{\sin \left[x\right]}{\left[x\right]},& \text{ if }& x\in (-\mathrm{\infty },0)\cup [1,\mathrm{\infty })\\ 0,& \text{ if }& x\in [0,1)\end{array}\right.$

$\Rightarrow \underset{x\to 0^{-}}{lim} f\left(x\right)=\underset{h\to 0^{-}}{lim} \frac{\sin [-h]}{[-h]}$

$$\begin{gathered} = \frac{\sin (-1)}{(-1)}=\sin 1 \\ \Rightarrow \underset{x\to 0^{+}}{lim} f\left(x\right)= \frac{\sin \left(0\right)}{\left(0\right)}=0 \end{gathered}$$

$\Rightarrow \underset{x\to 0^{-}}{lim} f\left(x\right)\ne \underset{x\to 0^{+}}{lim} f\left(x\right).$

therefore, $\underset{x\to 0}{lim} f\left(x\right)$ does not exist.