A) \[0\]
B) \[-1\]
C) \[1\]
D) 2
Correct Answer: C
Solution :
\[\underset{x\to -0}{\mathop{\lim }}\,\,{{(sin\,x)}^{1/x}}+\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{1}{x} \right)}^{\sin x}}\] \[=0+\underset{x\to 0}{\mathop{\lim }}\,e{{\,}^{\log {{\left( \frac{1}{x} \right)}^{\sin x}}}}\left\{ \begin{align} & as,\,{{(decimal)}^{\infty }}\to 0 \\ & \underset{x\to 0}{\mathop{\lim }}\,{{(\sin \,x)}^{1/x}}\to 0 \\ \end{align} \right\}\] \[={{e}^{\operatorname{l}\underset{x\to 0}{\mathop{im}}\,\frac{\log \left( \frac{1}{x} \right)}{\text{cosec}\,\text{x}},}}\] applying L-Hospital's rule, we get \[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x\left( -\frac{1}{{{x}^{2}}} \right)}{-\text{cosec}\,\,\text{x}\,\text{cot}\,\text{ x}}}}={{e}^{\underset{x\to 0}{\mathop{lim}}\,\,\,\frac{\sin x}{x}\,\tan x}}={{e}^{0}}=1\]
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