Question

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

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

Answer 1

The correct option is C $\frac{1}{2}$

$p=\underset{x\to 0+}{lim}(1+{\tan }^2\sqrt{x}{)}^{\frac{1}{2x}},1^{\infty }$ form

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

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

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

$=e^{\frac{1}{2}\cdot 1^2}=e^{\frac{1}{2}}$

Therefore $\ln p=\ln e^{1/2}=\frac{1}{2}$