Question

Evaluate the limit:
$\underset{x\to 0^{+}}{lim}\{1+{\tan }^2\sqrt{x}{\}}^{1/2x}$

Answer 1

Given:
$\underset{x\to 0^{+}}{lim}\{1+{\tan }^2\sqrt{x}{\}}^{1/2x}$

It becomes $1^{\infty }$ form

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,

$f\left(x\right)=1+{\tan }^2\sqrt{x}$

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

$=e^{\underset{x\to 0^{+}}{lim}\left(\frac{{\tan }^2\sqrt{x}}{2x}\right)}$

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

$=e^{1\times 1\times \frac{1}{2}}=\sqrt{e}$
$[\because \underset{x\to 0}{lim}\frac{\tan x}{x}=1]$