_ x 0 (1+x)^ 1 x - e x equals

MCQ

$\lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{x}}- e }{x}$ equals

A
$\frac{\pi}{2}$
B
$0$
C
$\frac{2}{ e }$
✓
$-\frac{ e }{2}$

✓

Answer

Correct option: D.

$-\frac{ e }{2}$

(D)
$(1+x)^{\frac{1}{x}}= e ^{\frac{1}{x}[\log (1+x)]}$
$= e ^{\frac{1}{x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots\right)}= e ^{\left(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots\right)}$
$=e \cdot e^{\left(-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots\right)}$
$= e \left[1+\left(-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots.\right)\right.$ $\left.+\frac{\left(-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}\right)}{2!}+\ldots.\right]$
$= e -\frac{ ex }{2}+\frac{1 le }{24} x^2-\ldots$
$\therefore \quad \lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{x}}- e }{x}=\lim _{x \rightarrow 0}\left(\frac{-\frac{ e x}{2}+\frac{11 e }{24} x^2-\ldots .}{x}\right)$
$=\lim _{x \rightarrow 0}\left(\frac{- e }{2}+\frac{11 e }{24} x-\ldots.\right)=-\frac{ e }{2}$

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