Question

If $f\left(x\right)={\left(\tan (\frac{\pi }{4}+\ln x)\right)}^{{\log }_{x}e}$ is to be made continuous at $x=1$, then $f\left(1\right)$ should be equal to

• A. $e^2$
• B. $e$
• C. $1/e$
• D. $e^{-2}$

Answer 1

The correct option is A $e^2$

For $f\left(x\right)$ to be continuous $x=1$

$f\left(1\right)={lim}_{x\to 1}{\left(\tan (\frac{\pi }{4}+\ln x)\right)}^{{\log }_{x}e}=1^{\infty }$ form

$=e^{{lim}_{x\to 1}\frac{1}{{\log }_{x}e}\left(\tan (\frac{\pi }{4}+\ln x)-1\right)}\frac{0}{0}$ form

$=e^{{lim}_{x\to 1}\frac{\left({\sec }^2(\pi /4+\ln x)\right)\times \left(1/x\right)+0}{\left(1/x\right)}}$

$=e^{{\sec }^2\pi /4}=e^2$