I need to solve the following: $$ \lim_{x \rightarrow 0} \frac{\sum_{n=1}^{\infty}(-1)^n \displaystyle \frac{x^{2n}}{(2n)!}}{\sum_{n=2}^{\infty} \displaystyle \frac{x^n}{n!}} $$
The answer given is $-1$.
Please advise on the initial steps to proceed, I am stuck because I know that $\sum(AB)\ne \sum(A)\sum(B)$ and $\sum\frac{A}{B}\ne\frac{\sum A}{\sum B}$, hence there is nothing much I can do with the expression.
\begin{align*} \lim_{x\rightarrow 0}\dfrac{\cos x-1}{e^{x}-x-1}=\lim_{x\rightarrow 0}\dfrac{-\sin x}{e^{x}-1}=\lim_{x\rightarrow 0}\dfrac{-\cos x}{e^{x}}=-1. \end{align*} Perhaps, \begin{align*} &\lim_{x\rightarrow 0}\dfrac{-x^{2}/2!+x^{4}/4!-\cdots}{x^{2}/2!+x^{3}/3!+\cdots}\\ &=\lim_{x\rightarrow 0}\dfrac{x^{-2}(-x^{2}/2!+x^{4}/4!-\cdots)}{x^{-2}(x^{2}/2!+x^{3}/3!+\cdots)}\\ &=\lim_{x\rightarrow 0}\dfrac{-1/2!+x^{2}/4!-\cdots}{1/2!+x/3!+\cdots}\\ &=\dfrac{-1/2}{1/2}\\ &=-1. \end{align*}