Functions Question 300
Question: The value of $ \underset{x\to 0}{\mathop{\lim }},\frac{\log
[1+x^{3}]}{{{\sin }^{3}}x}= $ [AMU 2005]
Options:
A) 0
B) 1
C) 3
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{x\to 0}{\mathop{\lim }},\frac{\log (1+x^{3})}{{{\sin }^{3}}x} $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{3x^{2}/(1+x^{3})}{3{{\sin }^{2}}x\cos x} $ [By using L- Hospital rule] $ =\underset{x\to 0}{\mathop{\lim }},[ \frac{1}{1+x^{3}}{{( \frac{x}{\sin x} )}^{2}}.\frac{1}{\cos x} ] $ $ =\frac{1}{1+0}.{{(1)}^{2}}.\frac{1}{1}=1 $ .
BETA
