Question

Let $p=\underset{x\to 0^{+}}{\lim }(1+{\tan }^2\sqrt{x}{)}^{\frac{1}{2x}}$ then $logp$ is equal to

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

Answer 1

Answer: (C)

$p=\underset{x\to 0^{+}}{\lim }{\left\{1+{\tan }^2\sqrt{x}\right\}}^{\frac{1}{{\tan }^2\sqrt{x}}\times \frac{{\tan }^2\sqrt{x}}{2x}}$
$=e^{\underset{x\to 0}{\lim }\frac{{\tan }^2\sqrt{x}}{{\left(\sqrt{x}\right)}^2}\times \frac{1}{2}}=e^{\frac{1}{2}}$
${\log }_{e}p=\frac{1}{2}$