Functions Question 293
Question: The value of $ \underset{x\to 0}{\mathop{\lim }},\frac{2}{x}\log (1+x) $ is equal to
[Pb. CET 2000]
Options:
A) e
B) $ e^{2} $
C) $ \frac{1}{2} $
D) 2
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{x\to 0}{\mathop{\lim }},\frac{2}{x}\log (1+x)=\underset{x\to 0}{\mathop{\lim }},2\log {{(1+x)}^{\frac{1}{x}}} $ $ =\underset{x\to 0}{\mathop{\lim }},2{\log_{e}}e=2 $ $ { \because \underset{x\to 0}{\mathop{\lim }},{{(1+x)}^{\frac{1}{x}}}={\log_{e}}e=1 } $ Trick : Using L Hospital?s rule.
BETA
