Question

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

Answer 1

Now,

$\underset{x\to 0}{lim}\frac{x{e}^x-\log (1+x)}{x^2}$ $\left(\frac{0}{0}\right)$ form.

Now applying L'Hospital's rule we get,

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

Now applying L'Hospital's rule we get,

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

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

$=\frac{3}{2}$.