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

# Let $P=\underset{x\to {0}^{+}}{lim} {\left(1+{\mathrm{tan}}^{2}\sqrt{x}\right)}^{2x}.$ Then , is equal to

1. A

1

2. B

$\frac{1}{2}$

3. C

$\frac{1}{4}$

4. D

2

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### Solution:

We have,

$\begin{array}{l}P=\underset{x\to {0}^{+}}{lim} {\left(1+{\mathrm{tan}}^{2}\sqrt{x}\right)}^{2x}\\ ⇒P={e}^{\underset{x\to {0}^{+}}{lim} 2x}1{\mathrm{lan}}^{2}\sqrt{x}={e}^{2}\underset{x\to {0}^{+}}{lim} {\left(\frac{\mathrm{tan}\sqrt{x}}{\sqrt{x}}\right)}^{2}\\ ⇒{\mathrm{log}}_{e}P=\frac{1}{2}\end{array}$

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