If $f\left(x\right)=\left\{\begin{array}{cl}\frac{\sin (1+[x\left]\right)}{\left[x\right]}& \text{ for }\left[x\right]\ne 0\\ 0& \text{ for }\left[x\right]=0\end{array}\right.$

denotes the greatest integer not exceeding $x,$ then $\underset{x\to 0^{-}}{lim} f\left(x\right)=$

Solution:

We have,

$\begin{array}{l}f\left(x\right)=\left\{\begin{array}{cl}\frac{\sin (1+[x\left]\right)}{\left[x\right]}& \text{ for }x\notin [0,1)\\ 0& \text{ for }x\in [0,1)\end{array}\right.\\ \therefore \underset{x\to 0^{-}}{lim} f\left(x\right)=\underset{x\to 0^{-}}{lim} \frac{\sin (1+[x\left]\right)}{\left[x\right]}=\underset{x\to 0^{-}}{lim} \frac{\sin (1-1)}{-1}=\frac{0}{-1}=0\end{array}$