Question

The value of $f\left(0\right)$ for $f\left(x\right)={(1+{\tan }^2\sqrt{x})}^{\frac{1}{2x}}$,so that $f\left(x\right)$ is continuous everywhere, is

• A. $e$
• B. $\frac{1}{2}$
• C. $e^{\frac{1}{2}}$
• D. $0$

Answer 1

Answer: (C)

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

Say $y=\sqrt{x}$.
For the function to be continuous at $x=0$, $f\left(0\right)={lim}_{x\to 0}f\left(x\right)$.
$\underset{y\to 0}{lim}{(1+{\tan }^{2}y)}^{\frac{1}{2y^2}}=\underset{y\to 0}{lim}{e}^{\frac{1}{2y^2}\log (1+{\tan }^{2}y)}=exp\left({lim}_{y\to 0}\frac{1}{2y^2}\log (1+{\tan }^{2}y)\right)=exp\left(\underset{y\to 0}{lim}\frac{{\tan }^{2}y}{2y^2}\frac{\log (1+{\tan }^{2}y)}{{\tan }^{2}y}\right)=exp(\frac{1}{2}\cdot 1)$
(Using $\underset{y\to 0}{lim}\frac{\tan y}{y}=1,\underset{y\to 0}{lim}\frac{\log (1+y)}{y}=1$)
This gives the limit, and hence the value of the function at $x=0$ as $e^{\frac{1}{2}}$.