[Math] using power expansion to find limit

Question

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!

Answer 1

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}

$(1), (2) \Rightarrow \displaystyle \lim_{x\to 0} \dfrac{x\cos x - \sin x}{x^2-x\log(1+x)} = \dfrac{-\dfrac{1}{3}}{\dfrac{1}{2}} = -\dfrac{2}{3}$.