Is this true: $\lim_{x\to 0} \frac{1-(\cos x)^n}{x^2}=\frac n2$ Does this statement hold $\forall n\in \Bbb R$?
$$\lim_{x\to 0} \frac{1-(\cos x)^n}{x^2}=\frac n2$$
I know that it holds for $\forall n \in \Bbb N$, because $1-(\cos x)^n=(1-\cos x)(1+ \cos x + ... + (\cos x)^{n-1})$, and $\lim_{x\to 0} (1+ \cos x + ... + (\cos x)^{n-1})=(1+1+...+1)=n$, but does that stand for, for example, $n=\sqrt2$?
If it does, how do I prove it?
 A: Note that for $n\in \mathbb{N}$
$$1-(\cos x)^n=(1-\cos x)\stackrel{\color{red}{\text{n terms}}}{(\cos^{n-1}x+\cos^{n-2}x+...+1)}$$
thus since $\cos x\to 1$
$$\frac{1-(\cos x)^n}{x^2}=\frac{(1-\cos x)}{x^2}\stackrel{\color{red}{\text{sum up to n}}}{(\cos^{n-1}x+\cos^{n-2}x+...+1)}\to\frac12\cdot n=\frac n2$$
For $n=a\in \mathbb{R}$ note that
$$\cos x=1-\frac{x^2}{2}+o(x^2) \quad (1+x)^a=1+ax+o(x) \\\implies \left(1-\frac{x^2}{2}+o(x^2)\right)^a=1-a\frac{x^2}{2}+o(x^2)$$
then
$$\frac{1-(\cos x)^n}{x^2}=\frac{1-1+a\frac{x^2}{2}+o(x^2)}{x^2}=\frac{a}2+o(1)\to\frac{a}2$$
A: $1-e^{a\cdot \ln(\cos(x))}=\frac{(a x^2)}{2} + o(x^2)$. 
So $\lim_{x\to 0 }\frac{1-\cos(x)^a}{x^2}=\lim_{x\to 0 }\frac{1-e^{a\cdot \ln(\cos(x))}}{x^2}=\frac{a}{2}$.
A: \begin{align}
\frac{1-(\cos x)^n}{x^2}&=\frac{1-\left(1-\frac{x^2}2+o(x^2)\right)^n}{x^2}\\
&=\frac{1-1+\frac{nx^2}2+o(x^2)}{x^2}\\
&=\frac{n}{2}+o(1)\\
\end{align}
A: The expression under limit can be written as $$\frac{\cos^{n}x-1}{\cos x-1}\frac{1-\cos x} {x^2}$$ Putting $t=\cos x$ we observe that $t\to 1$ so that the first factor $(t^n-1)/(t-1)\to n$ and the second factor tends to $1/2$. Thus the desired limit is $n/2$. One should use the standard limit formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ whenever applicable. 
A: Just a simple l’Hopitals rule:
$\lim_{x\to 0} \frac{1-(\cos x)^n}{x^2}=\lim_{x\to 0} \frac{(1-(\cos x)^n)'}{(x^2)'}=\lim_{x\to 0} \frac{0-(-n \sin x(\cos x)^{n-1})}{2x}=\lim_{x\to 0} \frac{n x(\cos x)^{n-1}}{2x}=\lim_{x\to 0}\frac{nx}{2x}=\frac n2$
