A) ?1
B) 0
C) 1
D) None of these
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{{{e}^{x}}-{{e}^{\sin x}}}{x-\sin x} \right]\], \[\left( \frac{0}{0}\text{form} \right)\] Using L-Hospital?s rule three times, then \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-{{e}^{\sin x}}.\cos x}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-{{e}^{\sin x}}{{\cos }^{2}}x+\sin x.{{e}^{\sin x}}}{\sin x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-{{e}^{\sin x}}.{{\cos }^{3}}x+{{e}^{\sin x}}2\cos x\sin x+{{e}^{\sin x}}.\cos x\sin x+{{e}^{\sin x}}.\cos x}{\cos x}\] \[=1\].
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