Question

${lim}_{x\to 0^{+}}{\{1+{\tan }^2\sqrt{x}\}}^{\frac{1}{2}x}$

Answer 1

${lim}_{x\to 0^{+}}{\{1+{\tan }^2\sqrt{x}\}}^{\frac{1}{2}x}=e^{{lim}_{x\to 0}\left\{\frac{{\tan }^2\sqrt{x}}{2x}\right\}}[\because {lim}_{x\to a}\{f\left(x\right){\}}^{g\left(x\right)}=e^{{lim}_{x\to a}f\left\{\right(x)-1\}g\left(x\right)}]=e^{{lim}_{x\to 0}\left\{\frac{si{n}^2\sqrt{x}}{2xco{s}^2\sqrt{x}}\right\}}=e^{\frac{1}{2}{lim}_{x\to 0^{+}}\left\{\left(\frac{si{n}^2\sqrt{x}}{\sqrt{x}}\right)\right\}{lim}_{x\to 0}\left\{\frac{1}{{\cos }^2\sqrt{x}}\right\}}=e^{\frac{1}{2}}=\sqrt{e}$