A) 0
B) 1
C) Does not exist
D) sin 1
Correct Answer: C
Solution :
[c] \[\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\sin [x-3]}{[x-3]} \right]\] For \[x\to {{0}^{+}},[x-3]=-3\] \[\therefore \frac{\sin [x-3]}{[x-3]}=\frac{\sin (-3)}{-3}=\frac{\sin 3}{3}\in (0,1)\] \[\therefore \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\sin [x-3]}{[x-3]}=0\] For \[x\to {{0}^{-}},[x-3]=-4\] \[\therefore \frac{\sin [x-3]}{[x-3]}=\frac{\sin 4}{4}\] lies in (-1, 0) \[\therefore \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \frac{\sin [x-3]}{[x-3]} \right]=-1\] \[\therefore \] Limit does not exist.
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