Functions Question 18
Question: $ \underset{x\to 0}{\mathop{\lim }}\frac{{e^{\sin x}}-1}{x}= $
Options:
A) 1
B) e
C) 1/e
D) None of these
Show Answer
Answer:
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. $
BETA
