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: (A)

$1$

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

as $x\to 0^{+}\frac{1}{x}\to \infty$

$={(1+0)}^{\infty }{1}^{\infty }$form

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

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

$\log P=\log e=1$