Question

Use series to evaluate the limit.
\[\lim_{x \rightarrow 0} \frac{ x - \ln (1+x) }{ x^2 }\]
ln (1+x) = (-1)^n-1 x^n/n
i tried using that and then dividing each term by x^2 ended up with
1/x-> 1/(1+(x-1)) --> (x-1)^n
and (-1)^n-1 x^n-2/n
but plugging in 0 doesn't get me anywhere here..