Let P=limx→0+(1+tan2x)1/2x . Then logeP is equal to

P=limx→0(1+tan2x)1/2xP=elimx→0+12x+tan2x=e1/2tan2x=e1/2limx→0+(tanxx)2=e1/2logep=12

Let P=limx→0+(1+tan2x)1/2x . Then logeP is equal to

P=limx→0(1+tan2x)1/2xP=elimx→0+12x+tan2x=e1/2tan2x=e1/2limx→0+(tanxx)2=e1/2logep=12

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Let $P = \lim_{x \to 0^{+}} {(1 + \tan^{2} \sqrt{x})}^{1 / 2 x}$ . Then $\log_{e}^{P}$ is equal to

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$2$

$1$

$\frac{1}{2}$

$\frac{1}{4}$

answer is B.

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$P = \lim_{x \to 0} {(1 + \tan^{2} \sqrt{x})}^{1 / 2 x}$

$P = e^{\lim_{x \to 0^{+}} \frac{1}{2 x} + \tan^{2} \sqrt{x}} = e^{1 / 2 \tan^{2} \sqrt{x}} = e^{1 / 2 \lim_{x \to 0^{+}} {(\frac{\tan \sqrt{x}}{\sqrt{x}})}^{2}} = e^{1 / 2}$

$\log_{e}^{p} = \frac{1}{2}$

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Let P=limx→0+(1+tan2x)1/2x . Then logeP is equal to

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Let $P = \lim_{x \to 0^{+}} {(1 + \tan^{2} \sqrt{x})}^{1 / 2 x}$ . Then $\log_{e}^{P}$ is equal to