Answer
$\frac{1}{2}$
Work Step by Step
$ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+....+\frac{x^{n}}{n}$
Plug into the limit to get
$\lim\limits_{x \to 0}\frac{x-ln(1+x)}{x^{2}}=\lim\limits_{x \to 0}x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+....+\frac{x^{n}}{n}$
$=\frac{1}{2}$
