Functions Question 14
Question: $ \underset{x\to 0}{\mathop{\lim }}\frac{2^{x}-1}{{{(1+x)}^{1/2}}-1}= $
[IIT 1983; Karnataka CET 1999]
Options:
A) $ \log 2 $
B) $ \log 4 $
C) $ \log \sqrt{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{x\to 0}{\mathop{\lim }}\frac{2^{x}-1}{{{(1+x)}^{1/2}}-1}=\underset{x\to 0}{\mathop{\lim }}\frac{2^{x}\log 2}{\tfrac{1}{2}{{(1+x)}^{-1/2}}} $ $ { \because \underset{x\to a}{\mathop{\lim }}\frac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\frac{{f}’(x)}{{g}’(x)} } $ $ =2\log 2=\log 4. $
BETA
