Question

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

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

Answer 1

Answer: (B)

$\frac{1}{2}$

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

$P=\underset{x\to 0^{+}}{lim}(1+{\tan }^2\sqrt{x}{)}^{\frac{1}{{\tan }^2\sqrt{x}}\times \frac{{\tan }^2\sqrt{x}}{2x}}$

$P=\underset{x\to 0^{+}}{lim}{e}^{\frac{{\tan }^2\sqrt{x}}{(\sqrt{x}{)}^2}\times \frac{1}{2}}=e^{\frac{1}{2}}$

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