First slide

Introduction to limits

Question

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

Moderate

A

$e^{1 / 2}$
B

$e^{1 / 4}$
C

$e^{1 / b}$
D

$e^{1 / 12}$

Solution

We have,
$\begin{array}{l} lim_{x \to 0} {(\frac{e^{x} + e^{- x} - 2}{x^{2}})}^{\frac{1}{x^{2}}} \\ = lim_{x \to 0} {\{\frac{2 (\frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + \dots)}{x^{2}}\}}^{\frac{1}{x^{2}}} \\ = lim_{x \to 0} {(1 + 2 \frac{x^{2}}{4!} + 2 \frac{x^{4}}{6!} + \dots)}^{\frac{1}{x^{2}}} \\ = lim_{x \to 0} (\frac{2 x^{2}}{4!} + \frac{2 x^{4}}{6!} + \dots) \times \frac{1}{x^{2}} = e^{1 / 12} \end{array}$

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