Evalutating $\lim_{x\to0}\frac{1-\cos x}{x^2}$ $$\lim_{x\to0}\frac{1-\cos x}{x^2}$$
I know there are many ways to calculate this. Like L'Hopital. But for learning purposes I am not supposed to do that. Instead, I decided to do it this way:
Consider that $\cos x = 1- \sin^2 \frac{x}{2}$ (from the doulbe-angle formulas here). Therefore:
$$\frac{1-(1- \sin^2 \frac{x}{2})}{x^2} = \frac{\sin^2 \frac{x}{2}}{x^2}$$
Let us split this:
$$\frac{\sin \frac{x}{2}}{x} \cdot \frac{\sin \frac{x}{2}}{x}$$
I will just work with the left one for now. At the end, I will just multiply it by itself.
$$\frac{\sin \frac{x}{2}}{x}$$
Hm. Remember that $\frac{\sin x}{x} = 1$ when $x\to0$. If only the denominator were an $\frac{x}{2}$ instead of $x$ I could do this... Well, I can! Let's just
$$\frac{\sin \frac{x}{2}}{x}\cdot\frac{\frac{1}{2}}{\frac{1}{2}}$$
Looks a bit dumb but it kind of makes sense. This will yield
$$\frac{\color{red}{\sin \frac{x}{2}}\cdot \frac{1}{2}}{\color{red}{\frac{x}{2}}}$$
Hurray, we can now apply the formula thingy and end up with
$$\frac{\color{red}1\cdot\frac{1}{2}}{\color{red}1} = \frac{1}{2}$$
Now I need to multiply this thing by itself.
$$\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$
And that is the answer.
... Which is wrong. The correct answer is
$$\frac{1}{2}$$
What did I do wrong in these steps? I know many of you are able to solve this with a variety of different methods, but I'm particularly interested in this specific procedure I just tried to use. What was my mistake?
 A: You are off by a factor of two in the numerator. $\cos x = 1- \color{red}2\sin^2 \frac{x}{2}$, whereas you put $\cos x = 1- \sin^2 \frac{x}{2}$
(There might be other mistakes, I'll continue checking)
A: $$\begin{align}
\lim_{x\to 0}\frac{1-\cos x}{x^2}\frac{1+\cos x}{1+\cos x} & = \lim_{x\to 0}\frac{\sin^2x}{x^2(1+\cos x)}\\
& = \lim_{x\to 0}\frac{\sin^2x}{x^2}\lim_{x\to 0}\frac 1{1+\cos x}\\
& = 1\cdot\frac 12=\frac 12.
\end{align}$$
A: Notice, your mistake  $\cos x\ne 1-\sin^2\frac{x}{2}$
Now, there are various methods to find the limit here given two methods as follows
Method-1$$\lim_{x\to 0}\frac{1-\cos x}{x^2}$$
$$=\lim_{x\to 0}\frac{2\sin^2\left( \frac{x}{2}\right)}{x^2}=\lim_{x\to 0}\frac{1}{2}\frac{\sin^2\left( \frac{x}{2}\right)}{\left( \frac{x}{2}\right)^2}$$$$=\frac{1}{2}\lim_{x\to 0}\left(\frac{\sin\left( \frac{x}{2}\right)}{\left( \frac{x}{2}\right)}\right)^2=\frac{1}{2}(1)^2=\frac{1}{2}$$
Method-2
$$\lim_{x\to 0}\frac{1-cos x}{x^2}$$
 Applying L'Hospital's rule for $\frac{0}{0}$ form  $$=\lim_{x\to 0}\frac{\sin x}{2x}=\frac{1}{2}\lim_{x\to 0}\frac{\sin x}{x}=\frac{1}{2}$$
