If $f(x)= \left\{ \begin{array}{1 1} \frac{\sin(1+[x])}{[x]}, & \quad for\;[x] \neq 0 \\ 0, & \quad for\;[x] =0 \end{array} \right.$ where $[x]$ denotes the greatest integer not exceeding x, then $\lim \limits _{x \to 0} f(x)$ is equal to

$\begin {array} {1 1} (a)\;-1 & \quad (b)\;0 \\ (c)\;1 & \quad (d)\;2 \end {array}$

(b) 0
answered Nov 7, 2013 by