A) \[1\]
B) \[0\]
C) \[-1\]
D) None of these
Correct Answer: D
Solution :
As, \[f(x)=\left\{ \begin{matrix} \frac{\sin [x]}{[x]}m & [x]\ne 0 \\ 0, & [x]=0 \\ \end{matrix} \right.\] \[\Rightarrow \] \[f(x)=\left\{ \begin{matrix} \frac{\sin [x]}{[x]}, & x\in R-[0,\,\,1) \\ 0, & 0\le x<1 \\ \end{matrix} \right.\] \[\therefore \] \[RHL\]at\[x=0\] \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin [0+h]}{[0+h]}=0\] \[LHL\]at\[x=0\] \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin [0-h]}{[0-h]}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin (-1)}{-1}=\sin 1\] Since, \[RHL\ne LHL\] \[\therefore \]Limit does not exist.
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