Functions Question 280
Question: $ \underset{x\to 0}{\mathop{\lim }}\frac{e^{x}-{e^{-x}}}{\sin x} $ is
[Kurukshetra CEE 2002]
Options:
A) 0
B) 1
C) 2
D) Non existent
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=\underset{x\to 0}{\mathop{\lim }},\frac{e^{x}-{e^{-x}}}{\sin x} $
Þ $ y=\underset{x\to 0}{\mathop{\lim }},\frac{[ 1+\frac{x}{1!}+\frac{x^{2}}{2!}+…. ]-[ 1-\frac{x}{1!}+\frac{x^{2}}{2!}-…. ]}{\sin x} $
Þ $ y=\underset{x\to 0}{\mathop{\lim }},\frac{2,[ \frac{x}{1!}+\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+…………. ]}{\sin x} $
Þ $ y=\underset{x\to 0}{\mathop{\lim }},\frac{2,[ 1+\frac{x^{2}}{3!}+\frac{x^{4}}{4!}+……….. ]}{\frac{\sin x}{x}} $
Þ $ y=\frac{\underset{x\to 0}{\mathop{\lim }},2,[ 1+\frac{x^{2}}{2!}+……. ]}{\underset{x\to 0}{\mathop{\lim }},\frac{\sin x}{x}} $
Þ $ y=\frac{2}{1}=2 $ Trick : Applying L-Hospital?s rule, $ \underset{x\to 0}{\mathop{\lim }},\frac{e^{x}-{e^{-x}}}{\sin x} $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{e^{x}+{e^{-x}}}{\cos x}=\frac{e^{0}+\frac{1}{e^{0}}}{\cos 0}=\frac{1+1}{1}=2 $ .
BETA
