A) -1
B) 1
C) does not exist
D) none of these
Correct Answer: C
Solution :
In closed interval of\[x=0\]at RHL \[[x]=0\] and at LHL also \[[0]=0\] \[\therefore \] \[f(x)=\frac{\sin [x]}{[x]}\] \[(-1\le x<0)\] \[=0\] \[(0\le x<1)\] \[\therefore \] \[LHL=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{\sin [x]}{[x]}\] \[=\frac{\sin (-1)}{-1}=\sin {{1}^{c}}\]and \[RHL=0\]hence limit does not existYou need to login to perform this action.
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