Functions Question 186
Question: The value of $ \underset{x\to 0}{\mathop{\lim }},\frac{{{(1+x)}^{1/x}}-e+\frac{1}{2}ex}{x^{2}} $ is
[DCE 2001]
Options:
A) $ \frac{11e}{24} $
B) $ \frac{-11e}{24} $
C) $ \frac{e}{24} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{(1+x)}^{1/x}}={e^{\frac{1}{x}\log ,(1+x)}}={e^{\frac{1}{x},( x-\frac{x^{2}}{2}+\frac{x^{3}}{3},-……. )}} $ $ ={e^{1-\frac{x}{2}+\frac{x^{2}}{3},-……}}=e,{e^{-,\frac{x}{2}+\frac{x^{2}}{3},-,…..}} $ $ =e,[ 1+( -\frac{x}{2}+\frac{x^{2}}{3}-….. )+\frac{1}{2!},{{( -\frac{x}{2}+\frac{x^{2}}{3},-,….. )}^{2}}+… ] $ $ =e,[ 1-\frac{x}{2}+\frac{11}{24}x^{2}-…. ] $
$ \therefore ,\underset{x\to 0}{\mathop{\lim }},\frac{{{(1+x)}^{1/x}}-e+\frac{ex}{2}}{x^{2}}=\frac{11e}{24} $ .
BETA
