Question

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

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

Answer 1

Answer: (C)

$\frac{1}{2}$

Answer 2

p = \displaystyle\lim_{x \to0^{+}} \left\\{ 1 + \tan^{2} \sqrt{x}\right\\}^{\frac{1}{\tan^{2} \sqrt{x}}\times\frac{\tan^{2}\sqrt{x}}{2x}}
$= e^{\displaystyle\lim_{x\to0} \frac{\tan^{2}\sqrt{x}}{\left(\sqrt{x}\right)^{2}} \times \frac{1}{2}} = e^{\frac{1}{2}}$
$\log_{e} p = \frac{1}{2}$