A) 0
B) 1
C) -1
D) Does not exist
Correct Answer: A
Solution :
[a] \[=\underset{x\to 0+}{\mathop{\lim }}\,\left[ \frac{\sin (sgn)x}{\sgn (x)} \right]=\underset{x\to 0+}{\mathop{\lim }}\,\left[ \frac{\sin 1}{1} \right]=0\] \[=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \frac{\sin (sgnx)}{\sgn (x)} \right]=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left[ \frac{\sin (-1)}{-1} \right]\] \[=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,[sin1]\] Hence, the given limit is 0.
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