Functions Question 243
Question: $ \underset{x\to 0}{\mathop{\lim }}\frac{{{(1+x)}^{1/x}}-e}{x} $ equals
[UPSEAT 2001]
Options:
A) $ \pi /2 $
B) 0
C) $ 2/e $
D) ? $ e/2 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{(1+x)}^{\frac{1}{x}}}={e^{\frac{1}{x},[\log (1+x)]}} $ $ ={e^{\frac{1}{x},( x,-,\frac{x^{2}}{2},+,\frac{x^{3}}{3},-,\frac{x^{4}}{4},+…. )}} $ $ ={e^{( 1,-,\frac{x}{2},+,\frac{x^{2}}{3},-,\frac{x^{3}}{4},+,…. )}} $ $ =e.{e^{( ,-,\frac{x}{2},+,\frac{x^{2}}{3},-,\frac{x^{3}}{4}+…. )}} $ $ =e[ \frac{( -\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+… )}{1!}+\frac{{{( -\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+… )}^{2}}}{2!}+… ] $ $ =[ e-\frac{ex}{2}+\frac{11e}{24}x^{2}+…+… ] $ \ $ \underset{x\to 0}{\mathop{\lim }},\frac{{{(1+x)}^{1/x}}-e}{x} $ $ =\underset{x\to 0}{\mathop{\lim }}[ \frac{e-\frac{ex}{2}-\frac{11e}{24}x^{2}+…e}{x} ] $
Þ $ \underset{x\to 0}{\mathop{\lim }}( -\frac{e}{2}-\frac{11e}{24}x+… ) $ $ =-\frac{e}{2} $ .
BETA
