Differentiation Question 197
Question: $ \underset{x\to 0}{\mathop{\lim }}{{{ \frac{1+tanx}{1+\sin x} }}^{\cos ecx}} $ is equal to
Options:
A) $ \frac{1}{e} $
B) 1
C) $ e $
D) $ e^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Consider $ \underset{x\to 0}{\mathop{\lim }}{{{ \frac{1+\tan x}{1+\sin x} }}^{\cos ecx}} $
$ =\underset{x\to 0}{\mathop{\lim }}\frac{{{[ {{( 1+\frac{\sin x}{\cos x} )}^{\frac{\cos x}{\sin x}}} ]}^{1/\cos x}}}{{{(1+sinx)}^{1/\sin x}}} $ We know, $ \underset{n\to 0}{\mathop{\lim }}( 1+{{\frac{1}{n}}^{n}} )=e $
$ \therefore \underset{x\to 0}{\mathop{\lim }}\frac{{{[ {{( 1+\frac{\sin x}{\cos x} )}^{\frac{\cos x}{\sin x}}} ]}^{1/\cos x}}}{{{(1+sinx)}^{1/\sin x}}} $
$ =\underset{x\to 0}{\mathop{\lim }}\frac{{{[ {{( 1+\frac{1}{\frac{\cos x}{\sin x}} )}^{\frac{\cos x}{\sin x}}} ]}^{\frac{1}{\cos x}}}}{[ {{( 1+\frac{1}{\cos ecx} )}^{\cos ecx}} ]} $
$ =\frac{{e^{\underset{x\to 0}{\mathop{\lim }}\frac{1}{\cos x}}}}{e}=\frac{e}{e}=1. $
BETA
