If $f(x)=\left\{\begin{matrix}\frac{\sin(1+[x])}{[x]}&;&for\,[x]≠0\\0&;&for\,[x]=0\end{matrix}\right.$, where [x] denotes the greatest integer ≤ x, then $\underset{x→0^-}{\lim}f(x)$ equals

1 0 -1 none of these

0

For −1 < x < 0, [x] = −1; so, $\underset{x→0^-}{\lim}\frac{\sin(1+[x])}{[x]}=\frac{\sin 0}{-1}=0$