Evaluate the following limit: _ x 0^+ 1+ ^2 x ^ 1 2 x - Vidyadip

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0^+}\Big\{1+\tan^2\sqrt{\text{x}}\Big\}^{\frac{1}{2}\text{x}}$

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Answer

$\lim\limits_{\text{x}\rightarrow0^+}\Big\{1+\tan^2\sqrt{\text{x}}\Big\}^{\frac{1}{2}\text{x}}$
$=\ \text{e}^{{\lim\limits_{\text{x}\rightarrow0^+}}\Big\{\frac{\tan^2\sqrt{\text{x}}}{2\text{x}}\Big\}}$
$=\ \text{e}^{{\lim\limits_{\text{x}\rightarrow0^+}}\Big\{\frac{\tan^2\sqrt{\text{x}}}{2\text{x}}\Big\}}$
$=\ \text{e}^{{\lim\limits_{\text{x}\rightarrow0^+}}\Big\{\frac{\sin^2\sqrt{\text{x}}}{2\text{x}\cos^2\sqrt{\text{x}}}\Big\}}$
$=\ \text{e}^{\frac{1}{2}\lim\limits_{\text{x}\rightarrow0^+}\bigg\{\Big(\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\Big)^2\bigg\}\lim\limits_{\text{x}\rightarrow0^+}\Big\{\frac{1}{\cos^2\sqrt{\text{x}}}\Big\}}$
$=\ \text{e}^\frac{1}{2}$
$=\ \sqrt{\text{e}}$

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