Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{a^x+a^{-x}-2}{x^2}$

Answer 1

Given:
$\underset{x\to 0}{lim}\frac{a^x+a^{-x}-2}{x^2}$

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

$=\underset{x\to 0}{lim}\frac{{(a^{\frac{x}{2}}-a^{-\frac{x}{2}})}^2}{x^2}$

$[\because (a-b{)}^2=a^2+b^2-2ab]$

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

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

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

$=\frac{(\log a{)}^2}{a^0}=(\log a{)}^2$
​​​​​​​$[\because \underset{x\to 0}{lim}{\left(\frac{a^x-1}{x}\right)}^2=\log a]$

Therefore,
$\underset{x\to 0}{lim}\frac{a^x+a^{-x}-2}{x^2}=(\log a{)}^2$