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$\lim\limits_{x\to 0}\large\frac{(1+x)^{1/x}-e}{x}$ is

$(a)\;1\qquad(b)\;\large\frac{e}{2}$$\qquad(c)\;\large\frac{-e}{2}$$\qquad(d)\;\large\frac{2}{e}$

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1 Answer

$(1+x)^{1/x}$
$\Rightarrow e^{\log(1+x)^{1/x}}=e^{\Large\frac{1}{x}\normalsize \log(1+x)}$
$\Rightarrow e^{\Large\frac{1}{x}(x-\frac{x^2}{2}+\frac{x^3}{3}......}$
$\Rightarrow e^{1-\Large\frac{x}{2}+\frac{x^2}{3}+.......}=e.e^{-\Large\frac{x}{2}+\frac{x^2}{3}...}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{(1+x)^{1/x}-c}{x}=$$\lim\limits_{x\to 0}\large\frac{e.e^{-\Large\frac{x}{2}+\frac{x^2}{3}-......}-e}{x}$
$\Rightarrow e^{\Large\lim\limits_{x\to 0}\large\frac{e^{-\Large\frac{x}{2}+\frac{x^2}{3}.......-1}}{-\Large\frac{x}{2}(2)}}$
$\Rightarrow e.\large\frac{1}{-2}$
$\Rightarrow \large\frac{-e}{2}$
Hence (c) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 
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