How to find the limit of $\lim_{x\to 0} \frac{1-\cos^n x}{x^2}$ How can I show that 

$$
\lim_{x\to 0} \frac{1-\cos^n x}{x^2} = \frac{n}{2}
$$

without using Taylor series $\cos^n x = 1 - \frac{n}{2} x^2 + \cdots\,$?

You just need to know that $\lim_{x\to 0}\frac{\sin x}{x}=1$, since:
$$\begin{eqnarray*} \frac{1-\cos^n x}{x^2} &=& \frac{1-\cos x}{x^2}\cdot \sum_{k=0}^{n-1}\cos^k(x) = \frac{2\sin^2\frac{x}{2}}{x^2}\cdot \sum_{k=0}^{n-1}\cos^k(x)\\ &=& \frac{1}{2}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \cdot \sum_{k=0}^{n-1}\cos^k(x)\stackrel{x\to 0}{\longrightarrow}\color{red}{\frac{n}{2}}.\end{eqnarray*}$$
