Question

If $f\left(x\right)=\frac{(-e^x+2^x)}{x}$ is continuous at $x=0$, then the value of $10|f\left(0\right)-{\log }_{e}2|$ is

Answer 1

For continuity of $f\left(x\right)$ at $x=0$, we must have,
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)$
$=\underset{x\to 0}{lim}\frac{(-e^x+2^x)}{x}\left[\frac{0}{0}\text{Form}\right]$
Applying L Hospital's rule
$=\underset{x\to 0}{lim}\frac{-e^x+2^x\ln 2}{1}=-e^0+2^0{\log }_{e}2$
$=-1+{\log }_{e}2$
$\therefore 10|f\left(0\right)-{\log }_{e}2|=10$