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Let $p=\lim_{x \rightarrow 0}(1+\tan^{2}\sqrt{x})^{\frac{1}{2x}}$, then  $\log p$ is equal to 

Answer:

Explanation:

Given $p=\lim_{x \rightarrow 0}(1+\tan^{2}\sqrt{x})^{\frac{1}{2x}}$ ( 1^∞ form)

$e^{\lim_{x \rightarrow0}\frac{\tan^{2}\sqrt{x}}{2x}}=e^{\frac{1}{2}\lim_{x \rightarrow 0}\left(\frac{\tan\sqrt{x}}{\sqrt{x}}\right)^{2}}=e^{\frac{1}{2}}$

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