A) -1
B) 0
C) 1
D) None of the above
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{{{e}^{\sin x}}-{{e}^{x}}}{\sin x-x} \right]=\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{{{e}^{x}}({{e}^{\sin -x-x}}-1)}{\sin x-x} \right]\] \[=\underset{x\to 0}{\mathop{\lim }}\,{{e}^{x}}\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{{{e}^{\sin x-x}}-1}{\sin x-x} \right]\] \[={{e}^{o}}\times 1=1\]
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