<s>[INST] 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 $\cos^n x = 1 - \frac{n}{2} x^2 + \cdots\,$?
 [/INST] I have seen several limit problems on MSE where people don't use the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ whereas frequent use is made of other standard limits like $$\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = \lim_{x \to 0}\frac{e^{x} - 1}{x} = 1\tag{2}$$ and this question is also an instance where the limit $(1)$ should be used.
We have
\begin{align}
L &= \lim_{x \to 0}\frac{1 - \cos^{n}x}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{1 - \cos^{n}x}{1 - \cos x}\cdot\frac{1 - \cos x}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{1 - \cos^{n}x}{1 - \cos x}\cdot\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\notag\\
&= \lim_{t \to 1}\frac{1 - t^{n}}{1 - t}\cdot\lim_{x \to 0}\frac{1 - \cos^{2}x}{x^{2}(1 + \cos x)}\text{ (putting }t = \cos x)\notag\\
&= \lim_{t \to 1}\frac{t^{n} - 1}{t - 1}\cdot\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\frac{1}{1 + \cos x}\notag\\
&= n\cdot 1\cdot\frac{1}{1 + 1}\notag\\
&= \frac{n}{2}\notag
\end{align}
</s>