Question

Solve the limit.
$\lim \limits_{x \to 0} \dfrac{ {e}^{x} + {e}^{ - x} - 2 }{ {x}^{2} }$
[No L'Hôpital's rule allowed]​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm \: \displaystyle\lim_{x \to 0}\rm \: \frac{{e}^{x} + {e}^{ - x} - 2}{ {x}^{2} }$

If we substitute directly x = 0, we get

$\rm \: = \: \frac{{e}^{0} + {e}^{ - 0} - 2}{ {0}^{2} }$

$\rm \: = \: \dfrac{1 + 1 - 2}{0}$

$\rm \: = \: \dfrac{2 - 2}{0}$

$\rm \: = \: \dfrac{0}{0}$

which is indeterminant form.

Consider, again

$\rm \: \displaystyle\lim_{x \to 0}\rm \: \frac{{e}^{x} + {e}^{ - x} - 2}{ {x}^{2} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \: \frac{{e}^{x} + \dfrac{1}{{e}^{x}} - 2}{ {x}^{2} }$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \: \frac{{e}^{2x} + 1 - 2{e}^{x}}{ {x}^{2} {e}^{x}}$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \: \frac{ {({e}^{x} - 1)}^{2} }{ {x}^{2} } \times \displaystyle\lim_{x \to 0}\rm {e}^{x}$

$\: \: \: \: \: \: \{ \: \because \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \: \}$

$\rm \: = \: \displaystyle\lim_{x \to 0}\rm \frac{{e}^{x} - 1}{x} \times \displaystyle\lim_{x \to 0}\rm \frac{{e}^{x} - 1}{x} \times {e}^{0}$

$\rm \: = \: 1 \times 1 \times 1$

$\rm \: = \: 1$

Hence,

$\rm\implies \: \: \boxed{\tt{ \:\rm \: \displaystyle\lim_{x \to 0}\rm \: \frac{{e}^{x} + {e}^{ - x} - 2}{ {x}^{2} } = 1 \: }}$

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ADDITIONAL INFORMATION

$\displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} \: = \: 1 \:$

$\displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} \: = \: 1 \:$

$\displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} \: = \: 1 \:$

$\displaystyle\lim_{x \to 0}\rm \frac{{e}^{x} - 1}{x} \: = \: 1 \:$

$\displaystyle\lim_{x \to 0}\rm \frac{{a}^{x} - 1}{x} \: = \: loga \:$