Question

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

Answer 1

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }\left[1+{\tan }^2\sqrt{x}\right]{}^{\frac{1}{2x}} \\ \mathrm{Using}\mathrm{the}\mathrm{theorem}\mathrm{given}\mathrm{below}: \\ \mathrm{If}\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }g\left(x\right)=0\mathrm{such}\mathrm{that}\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}\mathrm{exists},\mathrm{then}\underset{x\to a}{\lim }{\left[1+f\left(x\right)\right]}^{\frac{1}{g\left(x\right)}}=e^{\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}}. \\ \mathrm{Here}: \\ f\left(x\right)={\tan }^2\sqrt{x} \\ g\left(x\right)=2x \\ \Rightarrow e^{\underset{x\to 0^{+}}{\lim }\left(\frac{{\tan }^2\sqrt{x}}{2x}\right)} \\ =e^{\underset{x\to 0^{+}}{\lim }\left(\frac{\tan \sqrt{x}}{\sqrt{x}}\right)\times \left(\frac{\tan \sqrt{x}}{\sqrt{x}}\right)\times \frac{1}{2}} \\ =e^{1\times 1\times \frac{1}{2}} \\ =\sqrt{e} \end{gathered}$$