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

$\begin{array}{l}\underset{x\to 0^{-}}{Lt}f\left(x\right)=\underset{h\to 0}{Lt}\frac{\sin (-h)}{[-h]}=\underset{h\to 0}{Lt}\frac{\sin (-1)}{(-1)}=\sin 1,\text{\hspace{0.17em}where\hspace{0.17em}\hspace{0.17em}}x=0-h\\ \underset{x\to 0+}{Lt}f\left(x\right)=\underset{h\to 0}{Lt}\frac{\sin \left(h\right)}{\left[h\right]}=\frac{0}{0}form\\ \underset{x\to 0-}{Lt}f\left(x\right)\ne \underset{x\to 0+}{Lt}f\left(x\right)\end{array}$

$\underset{x\to 0}{Lt}f\left(x\right)$ does not exist