A) ?1
B) 1
C) 0
D) None of these
Correct Answer: D
Solution :
In closed interval of x = 0 at right hand side [x] = 0 and at left hand side \[[x]=-1.\] Also [0]=0. Therefore function is defined as \[f(x)=\left\{ \begin{align} & \frac{\sin \,[x]}{[x]}\,\,(-1\le x<0) \\ & \ \ \ \ \ 0\ \ (0\le x<1) \\ \end{align} \right.\] \Left hand limit \[=\underset{x\to 0-}{\mathop{\lim }}\,f(x)=\underset{x\to 0-}{\mathop{\lim }}\,\,\frac{\sin \,[x]}{[x]}\] \[=\frac{\sin \,(-1)}{-1}=\sin {{1}^{c}}\] Right hand limit = 0. Hence limit doesn't exist.
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