A) ?1
B) \[{{\log }_{e}}1\]
C) 1
D) None of these
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,x\log \sin x=\underset{x\to 0}{\mathop{\lim }}\,\,\log \,{{(\sin x)}^{x}}=\log \,[\underset{x\to 0}{\mathop{\lim }}\,\,\,{{(\sin x)}^{x}}]\] \[=\log \,\left[ \underset{x\to 0}{\mathop{\lim }}\,\,{{(1+\sin x-1)}^{\frac{x(\sin x-1)}{\sin x-1}}} \right]\] \[={{\log }_{e}}[{{e}^{\underset{x\to 0}{\mathop{\lim }}\,\,x(\sin x-1)}}]={{\log }_{e}}1.\]
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