Let P=limx→0+ 1+tan2⁡x2x. Then loge P, is equal to

A
1
B
$\frac{1}{2}$
C
$\frac{1}{4}$
D
2

Solution:

We have,

$\begin{array}{l} P = lim_{x \to 0^{+}} {(1 + \tan^{2} \sqrt{x})}^{2 x} \\ \Rightarrow P = e^{lim_{x \to 0^{+}} 2 x} 1 {lan}^{2} \sqrt{x} = e^{2} lim_{x \to 0^{+}} {(\frac{\tan \sqrt{x}}{\sqrt{x}})}^{2} \\ \Rightarrow \log_{e} P = \frac{1}{2} \end{array}$

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