Question

The value of $\underset{x\to 0}{\lim }\frac{xcosx-log\left(1+x\right)}{x^2}$ is

Answer 1

Answer: 0.5

$\underset{x\to 0}{\lim }\frac{xcosx-log\left(1+x\right)}{x^2}$ $(\frac{0}{0}$ Form)
Using L-Hospital's rule,
$\underset{x\to 0}{\lim }\frac{cosx-xsinx-\frac{1}{x+1}}{2x}$, $(\frac{0}{0}$ Form)
Again using L-Hospital's rule,
$=\underset{x\to 0}{\lim }\frac{-sinx-sinx-xcosx+\frac{1}{(x+1{)}^2}}{2}$
$=\frac{1}{2}=0.5$