The value of limx→0 1+tan2⁡x1/2x

A
$e$
B
$e^{1 / 2}$
C
$e^{1 / 4}$
D
$e^{- 1 / 2}$

Solution:

${(1 + \tan^{2} \sqrt{x})}^{\frac{1}{2 x}} = (1 + \tan^{2} \sqrt{x}) (\frac{1}{\tan^{2} \sqrt{x}}) \cdot {(\frac{\tan \sqrt{x}}{\sqrt{x}})}^{2} \frac{1}{2}$

$\to e^{1 / 2}$ as $x \to 0$ .

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