Functions Question 231
Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{x\tan 2x-2x\tan x}{{{(1-\cos 2x)}^{2}}} $ is
[IIT 1999]
Options:
A) 2
B) ?2
C) $ \frac{1}{2} $
D) $ -\frac{1}{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{x\to 0}{\mathop{\lim }}\frac{x\tan 2x-2x\tan x}{{{(1-\cos 2x)}^{2}}} $ $ =\underset{x\to 0}{\mathop{\lim }}\frac{x(\tan 2x-2\tan x)}{{{(2,{{\sin }^{2}}x)}^{2}}}=\underset{x\to 0}{\mathop{\lim }},\frac{1}{4},\frac{x,(\tan 2x-2\tan x)}{{{\sin }^{4}}x} $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{1}{4}\frac{x{ ( 2x+\frac{1}{3}{{(2x)}^{3}}+\frac{2}{15},{{(2x)}^{5}}+… )-2( x+\frac{x^{3}}{3}+\frac{2}{15}x^{5}+… ) }}{x^{4},{{( 1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}+…. )}^{4}}} $ $ =\frac{1}{4},.,( \frac{8}{3}-\frac{2}{3} )=\frac{2}{4}=\frac{1}{2} $ .
BETA
