using power expansion to find limit I am preparing for my final exam, and stuck on this question. 

Using power series expansion, evaluate $$\lim_{x\to 0} \frac{x\cos(x)
 -\sin(x)}{x^2-x\ln(1+x)}$$

I have no idea how to proceed. Any help would be highly appreciated!

We have: $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} +\cdots $,
$\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} +\cdots $,
$\log(1+x) = x -\dfrac{x^2}{2} +\dfrac{x^3}{3} + \cdots $
Thus:
\begin{align}
x\cos x - \sin x 
  &= \left(x - \dfrac{x^3}{2!} + \dfrac{x^5}{4!}+\cdots \right) - \left(x -   
    \dfrac{x^3}{3!} + \dfrac{x^5}{5!} + \cdots\right)        \\
  &= x^3\left(-\dfrac{1}{3} + \dfrac{x^2}{30}+ O(x^4)\right) \\
  &= x^3\left(-\dfrac{1}{3} + O(x^2)\right)\tag{1}
\end{align}
\begin{align}
x^2 - x\log (1+x) 
  &= x^2 - x\left(x - \dfrac{x^2}{2} +\dfrac{x^3}{3}+\cdots \right) \\
  &= \dfrac{x^3}{2} - \dfrac{x^4}{3} + \dfrac{x^5}{4} + \cdots      \\
  &= x^3\left(\dfrac{1}{2} - \dfrac{x}{3}+\dfrac{x^2}{4}\right)     \\
  &= x^3\left(\dfrac{1}{2} + O(x)\right)\tag{2}.
\end{align}
$(1), (2) \Rightarrow \displaystyle \lim_{x\to 0} \dfrac{x\cos x - \sin x}{x^2-x\log(1+x)} = \dfrac{-\dfrac{1}{3}}{\dfrac{1}{2}} = -\dfrac{2}{3}$.
