A) \[\frac{11e}{24}\]
B) \[\frac{-11e}{24}\]
C) \[\frac{e}{24}\]
D) None of these
Correct Answer: A
Solution :
\[{{(1+x)}^{1/x}}={{e}^{\frac{1}{x}\log \,(1+x)}}={{e}^{\frac{1}{x}\,\left( x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}\,-....... \right)}}\] \[={{e}^{1-\frac{x}{2}+\frac{{{x}^{2}}}{3}\,-......}}=e\,{{e}^{-\,\frac{x}{2}+\frac{{{x}^{2}}}{3}\,-\,.....}}\] \[=e\,\left[ 1+\left( -\frac{x}{2}+\frac{{{x}^{2}}}{3}-..... \right)+\frac{1}{2\,\,!}\,{{\left( -\frac{x}{2}+\frac{{{x}^{2}}}{3}\,-\,..... \right)}^{2}}+... \right]\] \[=e\,\left[ 1-\frac{x}{2}+\frac{11}{24}{{x}^{2}}-.... \right]\] \[\therefore \,\,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{(1+x)}^{1/x}}-e+\frac{ex}{2}}{{{x}^{2}}}=\frac{11e}{24}\].
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