Functions Question 289
Question: $ \underset{x\to 0}{\mathop{\lim }},
[ \frac{e^{x}-{e^{\sin x}}}{x-\sin x} ] $ is equal to [UPSEAT 2004]
Options:
A) ?1
B) 0
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{x\to 0}{\mathop{\lim }},[ \frac{e^{x}-{e^{\sin x}}}{x-\sin x} ] $ , $ ( \frac{0}{0}form ) $ 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 $ .
BETA
