Functions Question 302
Question: $ \underset{\theta \to 0}{\mathop{\lim }},\frac{4\theta (\tan \theta -2\theta \tan \theta )}{{{(1-\cos 2\theta )}^{2}}} $ is
[Orissa JEE 2005]
Options:
A) $ 1/\sqrt{2} $
B) 1/2
C) 1
D) 2
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{\theta \to 0}{\mathop{\lim }},\frac{4\theta (\tan \theta -\sin \theta )}{{{(1-\cos 2\theta )}^{2}}}=\underset{\theta \to 0}{\mathop{\lim }},\frac{4\theta \sin \theta (1-\cos \theta )}{4{{\sin }^{4}}\theta \cos \theta } $ $ =\underset{\theta \to 0}{\mathop{\lim }},( \frac{\theta }{\sin \theta } )\frac{2{{\sin }^{2}}\theta /2}{{{\sin }^{2}}\theta \cos \theta } $ $ =\underset{\theta \to 0}{\mathop{\lim }},\frac{2{{\sin }^{2}}\theta /2}{(2\sin (\theta /2)\cos {{(\theta /2)}^{2}})}\frac{1}{\cos \theta } $ $ =\underset{\theta \to 0}{\mathop{\lim }},\frac{1}{2}\frac{1}{{{\cos }^{2}}(\theta /2).\cos \theta }=\frac{1}{2} $ .
BETA
