evaluating limit of $\lim_{x \to 0}\frac{1-\cos(x)}{\sin(x)(e^x-1)}$ This is for evaluating limit of $\lim_{x \to 0}\frac{1-\cos(x)}{\sin(x)(e^x-1)}$. It's easy with evaluating using L'Hôpital's Rule, but I want to use Taylor series. I can see here $\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$, $\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$, also $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$. So I use these to find it. What to do next?

Let $\mathcal{L}$ be your limit :
\begin{align}
\mathcal{L}&=\lim_{x \to 0}\frac{1-\cos x}{\sin x(e^x-1)}\\
&=\lim_{x\to 0} \frac{1-\cos x}{x^2}\frac{x^2}{\sin x (e^x-1)}\\
&=\lim_{x\to 0} \frac{1-\cos x}{x^2}\frac{x}{\sin x} \frac{x}{e^x-1}
\end{align}
Recall :
$$\lim_{x\to 0} \frac{1-\cos x}{x^2} = \frac{1}2  \ \ \ \ \ ;\ \ \ \ \ \lim_{x\to 0} \frac{\sin x}{x}=1\ \ \ \ \ ;\ \ \ \ \ \lim_{x\to 0}\frac{e^x-1}{x}=1 $$
Using these usual limits, you can easily get :
$$\mathcal{L}=\frac{1}{2}$$
