Functions Question 103
Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{\tan x-\sin x}{x^{3}}= $
[IIT 1974; AI CBSE 1986, 90; AISSE 1983, 86, 90; RPET 2000]
Options:
A) $ \frac{1}{2} $
B) $ -\frac{1}{2} $
C) $ \frac{2}{3} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{x\to 0}{\mathop{\lim }}\frac{\tan x-\sin x}{x^{3}}=\underset{x\to 0}{\mathop{\lim }}\frac{\sin x-\sin x,\cos x}{x^{3}\cos x} $ $ =\underset{x\to 0}{\mathop{\lim }}\frac{\sin x,( 2{{\sin }^{2}}\frac{x}{2} )}{x^{3},\cos x}=\underset{x\to 0}{\mathop{\lim }}[ \frac{\sin x}{x}.\frac{2}{\cos x}.\frac{{{\sin }^{2}}\frac{x}{2}}{{{( \frac{x}{2} )}^{2}}}.\frac{1}{4} ]=\frac{1}{2} $ .
BETA
