Question

If $L=\underset{x\to \infty }{\lim }\left(x-x^2{\log }_e\left(1+\frac{1}{x}\right)\right)$, then the value of $8L$ is

Answer 1

Answer: 4

Let $x=1/y$
$\Rightarrow \underset{x\to \infty }{\lim }\left(x-x^2{\log }_e\left(1+\frac{1}{x}\right)\right)$
$=\underset{y\to 0}{\lim }\left(\frac{1}{y}-\frac{{\log }_e(1+y)}{y^2}\right)$
$=\underset{y\to 0}{\lim }\left(\frac{y-{\log }_e(1+y)}{y^2}\right)$
$=\underset{y\to 0}{\lim }\left(\frac{y-\left(y-\frac{y^2}{2}\right)}{y^2}\right)=1/2$