A) 1
B) e
C) 1/e
D) None of these
Correct Answer: A
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{e}^{\sin x}}-1}{x}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{e}^{\sin x}}-1}{\sin x}\times \frac{\sin x}{x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{e}^{\sin x}}-1}{\sin x}\times \underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sin x}{x}=1\times 1=1\]. Aliter : Apply L-Hospital?s rule, \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{e}^{\sin x}}-1}{x}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\cos x\,{{e}^{\sin x}}}{1}=1.\,{{e}^{0}}=1.\]
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