A) \[\pi /2\]
B) 0
C) \[2/e\]
D) ?\[e/2\]
Correct Answer: D
Solution :
\[{{(1+x)}^{\frac{1}{x}}}={{e}^{\frac{1}{x}\,[\log (1+x)]}}\] \[={{e}^{\frac{1}{x}\,\left( x\,-\,\frac{{{x}^{2}}}{2}\,+\,\frac{{{x}^{3}}}{3}\,-\,\frac{{{x}^{4}}}{4}\,+.... \right)}}\]\[={{e}^{\left( 1\,-\,\frac{x}{2}\,+\,\frac{{{x}^{2}}}{3}\,-\,\frac{{{x}^{3}}}{4}\,+\,.... \right)}}\] \[=e.{{e}^{\left( \,-\,\frac{x}{2}\,+\,\frac{{{x}^{2}}}{3}\,-\,\frac{{{x}^{3}}}{4}+.... \right)}}\] \[=e\left[ \frac{\left( -\frac{x}{2}+\frac{{{x}^{2}}}{3}-\frac{{{x}^{3}}}{4}+... \right)}{1!}+\frac{{{\left( -\frac{x}{2}+\frac{{{x}^{2}}}{3}-\frac{{{x}^{3}}}{4}+... \right)}^{2}}}{2!}+... \right]\] \[=\left[ e-\frac{ex}{2}+\frac{11e}{24}{{x}^{2}}+...+... \right]\] \\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{(1+x)}^{1/x}}-e}{x}\]\[=\underset{x\to 0}{\mathop{\lim }}\,\,\left[ \frac{e-\frac{ex}{2}-\frac{11e}{24}{{x}^{2}}+...e}{x} \right]\] Þ \[\underset{x\to 0}{\mathop{\lim }}\,\,\left( -\frac{e}{2}-\frac{11e}{24}x+... \right)\]\[=-\frac{e}{2}\].
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