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 $\cos^n x = 1 - \frac{n}{2} x^2 + \cdots\,$?
 A: From l'Hopital's rule ...
$$L = \lim_{x\to 0}\frac{1-\cos^n x}{x^2} = \lim_{x\to 0}\frac{n\sin x\cos^{n-1}x}{2x} = \lim_{x\to 0}\frac n2\left(\frac{\sin x}x\right)\left(\cos^{n-1}x\right) = \cdots$$
A: Here, we use an approach that is more efficient and more elementary than use of L'Hospital's Rule.
Simply factor the term $1-\cos^n(x)$ as
$$1-\cos^n(x)=(1-\cos(x))\sum_{m=0}^{n-1}\cos^m(x)$$
Then, we have
$$\begin{align}
\lim_{x\to 0}\frac{1-\cos^n(x)}{x^2}&=\lim_{x\to 0}\frac{1-\cos(x)}{x^2}\sum_{m=0}^{n-1}\cos^m(x)\\\\
&\lim_{x\to 0}\frac{1-\cos(x)}{x^2} \lim_{x\to 0} \sum_{m=0}^{n-1}\cos^m(x)\\\\
&=\left(\frac12\right)(n)\\\\
&=\frac n2
\end{align}$$
A: 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}
A: $$\lim_{x\to0}\frac{1-\cos^n(x)}{x^2}=$$

Applying l'Hôpital's rule, we get that:

$$\lim_{x\to0}\frac{\frac{\partial}{\partial x}\left(1-\cos^n(x)\right)}{\frac{\partial}{\partial x}\left(x^2\right)}=\lim_{x\to0}\frac{n\sin(x)\cos^{n-1}(x)}{2x}=$$

By the product rule:

$$\frac{1}{2}\left[\lim_{x\to0}\frac{n\sin(x)}{x}\right]\left[\lim_{x\to0}\cos^{n-1}(x)\right]=\frac{1}{2}\left[\lim_{x\to0}\frac{n\sin(x)}{x}\right]\left[\cos^{n-1}(0)\right]=$$
$$\frac{1}{2}\left[\lim_{x\to0}\frac{n\sin(x)}{x}\right]\left[1\right]=\frac{1}{2}\left[\lim_{x\to0}\frac{n\sin(x)}{x}\right]=\frac{1}{2}\left[n\lim_{x\to0}\frac{\sin(x)}{x}\right]=$$

Applying l'Hôpital's rule, we get that:

$$\frac{1}{2}\left[n\lim_{x\to0}\frac{\frac{\partial}{\partial x}\left(\sin(x)\right)}{\frac{\partial}{\partial x}\left(x\right)}\right]=\frac{1}{2}\left[n\lim_{x\to0}\frac{\cos(x)}{1}\right]=\frac{1}{2}\left[n\lim_{x\to0}\cos(x)\right]=$$
$$\frac{1}{2}\left[n\cos(0)\right]=\frac{1}{2}\left[n\cdot1\right]=\frac{1}{2}\left[n\right]=\frac{n}{2}$$
A: You just need to know that $\lim_{x\to 0}\frac{\sin x}{x}=1$, since:
$$\begin{eqnarray*} \frac{1-\cos^n x}{x^2} &=& \frac{1-\cos x}{x^2}\cdot \sum_{k=0}^{n-1}\cos^k(x) = \frac{2\sin^2\frac{x}{2}}{x^2}\cdot \sum_{k=0}^{n-1}\cos^k(x)\\ &=& \frac{1}{2}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \cdot \sum_{k=0}^{n-1}\cos^k(x)\stackrel{x\to 0}{\longrightarrow}\color{red}{\frac{n}{2}}.\end{eqnarray*}$$
A: Without L'Hospital:
Thanks to $\dfrac{\sin(x)}x\to1$, one can interchange $x$ and $\sin(x)$ and rewrite the limits as
$$\lim_{x\to0}\frac{1-(1-x^2)^{n/2}}{x^2}=\lim_{x\to0}\frac{1-(1-x^2)^n}{x^2(1+(1-x^2)^{n/2})}.$$
And by the binomial theorem,
$$
\lim_{x\to0}\frac{1-1+nx^2-\binom n2x^4+\binom n3x^6\cdots}{2x^2}=\frac n2.$$
