Calculating the limit of $\frac{\cos(x)-1}{x}$ as $x \rightarrow 0$ Show that the $\lim \limits_{x \to 0}\frac{\cos(x)-1}{x}=0$
My attempt 
$$\begin{align}
\lim \limits_{x \to 0}\frac{\cos(x)-1}{x}*\frac{\cos(x)+1}{\cos(x)+1}&=\lim \limits_{x \to 0}\frac{1-\cos^2(x)}{x(1+\cos(x))}\\\\
&=\lim \limits_{x \to 0}\frac{\sin^2(x)}{x(1+\cos(x))}\\\\
&=\lim \limits_{x \to 0}\frac{\sin(x)}{x}*\frac{\sin(x)}{1+\cos(x)}\\\\
&=1*\frac{0}{2}\\\\
&=0
\end{align}$$ $$QED$$
Since $\lim \limits_{x \to 0}\frac{\sin(x)}{x}=1$ by the sandwich theorem.
i know this is correct however i would like it if anyone could show me the natural argument of using power series instead.
Recall the definition of the cosine function by the power series:
$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n}, \forall x \in \mathbb R$$
Also can anyone show me how to show that the radius of convergence of the cosine power series is $$\infty$$
 A: If you insist in power series:
$$\cos x=1-\frac{x^2}2+\frac{x^4}{24}-\ldots\implies \cos x-1=x\left(-\frac x{2}+\frac{x^3}{24}-\ldots\right)\implies$$
$$\frac{\cos x-1}x=\left(-\frac x{2}+\frac{x^3}{24}-\ldots\right)=-\frac x2+\mathcal O(x^3)\xrightarrow[x\to0]{}0$$
A: Note that $\cos(x)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}$.  Therefore, we can write
$$\frac{\cos(x)-1}{x}=\sum_{n=1}^\infty\frac{(-1)^n x^{2n-1}}{(2n)!} \tag 1$$
Hence, inasmuch as the lowest order power in the series is $1$, all terms approach $0$ as $x\to 0$.
Interestingly, we immediately have from $(1)$ that 
$$\lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac12$$

Second Question:
The radius of convergence for the power series for the cosine function can be found using the ratio test.  Note that the series converges when
$$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|\le1$$
where $a_n=\frac{(-1)^nx^{2n}}{(2n)!}$.  Proceeding we see that
$$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to \infty}\frac{x^2}{(2n+2)(2n+1)}=0$$
for all $x$.  Hence, the radius of convergence is indeed $\infty$.

ASIDE:
Note that $1-\cos(x)=2\sin^2(x/2)$.  Therefore, we can write
$$\frac{\cos(x)-1}{x}=-\underbrace{\left(\frac{\sin(x/2)}{x/2}\right)}_{\to1}\,\underbrace{\sin(x/2)}_{\to0}$$
A: Hint:
using the series  of $\cos x$ you have:
$$
\cos x-1 =-1 +\cos x= -1+1-\frac{x^2}{2!}+o(x^3)= -\frac{x^2}{2}+o(x^3)
$$
