First slide

Introduction to limits

Question

The value of $lim_{x \to 0} \frac{(1 + x)^{1 / x} - e + \frac{1}{2} e x}{x^{2}}$ is

Moderate

A

$\frac{11 e}{24}$
B

$\frac{- 11 e}{24}$
C

$\frac{e}{24}$
D

none of these

Solution

We have,
$lim_{x \to 0} \frac{(1 + x)^{1 / x} - e + \frac{e x}{2}}{x^{2}}$
$= lim_{x \to 0} \frac{e^{\frac{1}{-} \log (1 + x)} - e + \frac{e x}{2}}{x}$
$= lim_{x \to 0} \frac{e^{1 - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + \dots} - e + \frac{e x}{2}}{x}$
$= e^{lim_{x \to 0}} \frac{e^{- \frac{x}{2} + \frac{x^{2}}{3}} \frac{x^{3}}{4} + \dots - 1 + \frac{x}{2}}{x^{2}}$
$= e lim_{x \to 0} \frac{1 + (- \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + \dots) + \frac{1}{2!} {(- \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + \dots)}^{2} + \dots - 1 + \frac{x}{2}}{x^{2}}$
$= e lim_{x \to 0} \frac{x^{2} (\frac{1}{3} + \frac{1}{2!} \times \frac{1}{4}) + x^{3} (\dots) + \dots}{x^{2}} = e (\frac{1}{3} + \frac{1}{8}) = \frac{11 e}{24}$

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