Question

$\underset{x\to 0}{lim}\frac{x{e}^x-\log (1+x)}{x^2}$ equals

Answer 1

$\underset{x\to 0}{lim}\frac{x{e}^x-\log (1+x)}{x^2}\Rightarrow \frac{f\left(x\right)}{g\left(x\right)}\Rightarrow \frac{f\left(0\right)}{g\left(0\right)}=\frac{0}{0}$

By L-Hopital rule

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}=\underset{x\to 0}{lim}\frac{e^x-\frac{1}{(1+x)}+x{e}^x}{2x}$

$=\underset{x\to 0}{lim}\frac{e^x+\frac{1}{(1+x{)}^2}+e^x+x{e}^x}{2}=\frac{e^0+\frac{1}{1^2}+e^0+0}{2}$

$=\frac{1+1+1}{2}=\frac{3}{2}$