Differentiation Question 161
Question: If $ m,n\in I_0 $ and $ \underset{x\to 0}{\mathop{\lim }}\frac{\tan 2x-n\sin x}{x^{3}}= $ some integer, then value of this limit is
Options:
A) 3
B) 2
C) $ \frac{16+n}{12} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \underset{x\to 0}{\mathop{\lim }}\frac{\tan 2x-n\sin x}{x^{3}}=I $
$ =\underset{x\to 0}{\mathop{\lim }}\frac{2x+\frac{8x^{3}}{3!}…-nx+\frac{nx^{3}}{3!}}{x^{3}}=I $
$ =\underset{x\to 0}{\mathop{\lim }}\frac{(2-n)x+( \frac{16+n}{6} )x^{3}+…}{x^{3}}=I $
$ \Rightarrow n=2 $ and, thus required value $ =\frac{16+n}{6}=3. $
BETA
