Functions Question 62
Question: $ \underset{x\to 0}{\mathop{\lim }},{ \frac{\sin x-x+\frac{x^{3}}{6}}{x^{5}} }= $
[MNR 1985]
Options:
A) 1/120
B) ?1/120
C) 1/20
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Expand $ \sin x $ and then solve. Aliter : Apply L-Hospital?s rule $ \underset{x\to 0}{\mathop{\lim }},\frac{\sin x-x+\frac{x^{3}}{6}}{x^{5}}=\underset{x\to 0}{\mathop{\lim }},\frac{\cos x-1+\frac{3x^{2}}{6}}{5x^{4}} $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{-\sin x+\frac{6x}{6}}{20x^{3}}=\underset{x\to 0}{\mathop{\lim }}\frac{-\cos x+1}{60x^{2}}=\underset{x\to 0}{\mathop{\lim }},\frac{\sin x}{120,x} $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{\cos x}{120}=\frac{1}{120}. $
BETA
