Question

Evaluate the limit:
$\underset{x\to 0}{lim}{\left\{\frac{e^x+e^{-x}-2}{x^2}\right\}}^{1/x^2}$

Answer 1

Given:
$\underset{x\to 0}{lim}{\left\{\frac{e^x+e^{-x}-2}{x^2}\right\}}^{1/x^2}$

$=\underset{x\to 0}{lim}{\left\{\frac{{\left(e^{\frac{x}{2}}\right)}^2+{\left(e^{-\frac{x}{2}}\right)}^2-2e^{\frac{x}{2}}{e}^{-\frac{x}{2}}}{x^2}\right\}}^{1/x^2}$

$=\underset{x\to 0}{lim}{\left\{\frac{{(e^{\frac{x}{2}}-e^{-\frac{x}{2}})}^2}{x^2}\right\}}^{1/x^2}$

$=\underset{x\to 0}{lim}{\left\{{\left(\frac{e^x-1}{e^{\frac{x}{2}}\cdot x}\right)}^2\right\}}^{1/x^2}$

$=\underset{x\to 0}{lim}{\{\frac{1}{e^x}\times {\left(\frac{e^x-1}{x}\right)}^2\}}^{1/x^2}$

$={\{\frac{1}{e^0}\times (1{)}^2\left)\right\}}^{1/0}=1^{\infty }$

$[\because \underset{x\to 0}{lim}\frac{a^x-1}{x}=1]$

$\underset{x\to 0}{lim}{\left\{\frac{e^x+e^{-x}-2}{x^2}\right\}}^{1/x^2}$ is of the form of $1^{\infty }$

Using the result below:

If $\underset{x\to a}{lim}f\left(x\right)=1$ and $\underset{x\to a}{lim}g\left(x\right)=\infty$ such that $\underset{x\to a}{lim}\left\{f\right(x)-1\}g\left(x\right)$ exists, then

$\underset{x\to a}{lim}\left\{f\right(x){\}}^{g\left(x\right)}=e^{\underset{x\to a}{lim}\left\{f\right(x)-1\}g\left(x\right)}$

Here,

$f\left(x\right)=\frac{e^x+e^{-x}-2}{x^2}$

$g\left(x\right)=\frac{1}{x^2}$

$=e^{\underset{x\to 0}{lim}}\left(\frac{e^x+e^{-x}-2}{x^2}\right)\times \left(\frac{1}{x^2}\right)$

$\because e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdot \cdot \cdot$

$e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdot \cdot \cdot$

$e^x+e^{-x}=2+\frac{2x^2}{2!}+\frac{2x^4}{4!}+\cdot \cdot \cdot$

$=e^{\underset{x\to 0}{lim}(\frac{(2+\frac{2x^2}{2!}+\frac{2x^4}{4!}+\cdot \cdot \cdot )-2}{x^2}-1)\times \left(\frac{1}{x^2}\right)}$

$=e^{\underset{x\to 0}{lim}\left(\frac{x^2+\frac{2x^4}{4!}+\cdot \cdot \cdot -x^2}{x^2}\right)\times \left(\frac{1}{x^2}\right)}$

$=e^{\underset{x\to 0}{lim}\left(\frac{\frac{2x^4}{4!}+\frac{2x^6}{6!}+\cdot \cdot \cdot }{x^4}\right)}$

$=e^{\underset{x\to 0}{lim}(\frac{2}{4!}+\frac{2x^2}{6!}+\cdot \cdot \cdot )}$

$=e^{\frac{2}{4!}}=e^{\frac{1}{12}}$

Therefore,
$\underset{x\to 0}{lim}{\left\{\frac{e^x+e^{-x}-2}{x^2}\right\}}^{1/x^2}=e^{\frac{1}{12}}$