How to find the given limits I now that: $$\lim_{x\to 0}\frac{\sin x}{x}=1,$$ but I didnt now how to prove that:$$\lim_{x\to 0}\frac{1-\cos x}{x^2}$$
Please help me. Thanky very much for your help.
 A: Hint:
Put $\cos x=1- 2\sin^2\frac{x}{2}$.
A: Notice that this is a $0/0$ indeterminant form, so you can also apply L'Hopital's rule:
$\displaystyle \lim_{x \rightarrow 0 } \frac{1 - \cos(x)}{x^2} = \lim_{x \rightarrow 0 } \frac{ \sin(x)}{2x}$, and limit arithmetic tells us that this is equal to $\displaystyle \frac{1}{2} \left( \lim_{x \rightarrow 0 } \frac{\sin(x)}{x} \right)$.
A: A simpler hint for the second limit:
$$\frac{1-\cos x}{x^2}=\frac{(1-\cos x)(1+\cos x)}{x^2(1+\cos x)}=\dotsm$$
A: First method:
$$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\left(\frac{1-\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x}\right)=\lim_{x\to 0}\frac{1-\cos^2 x}{x^2(1+\cos x)}$$
$$=\lim_{x\to 0}\frac{\sin^2 x}{x^2(1+\cos x)}=\lim_{x\to 0}\frac{\sin x\cdot \sin x}{x\cdot x (1+\cos x)}=\lim_{x\to 0}\left(\frac{\sin x}{x}\cdot\frac{\sin x}{x}\cdot \frac{1}{1+\cos x}\right)$$
$$=\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{1}{1+\cos x}=1\cdot 1\cdot \frac{1}{1+\lim_{x\to 0} \cos x}=\frac{1}{2}$$
Second method:
Use the trigonometrics formula, $1-\cos x=2\sin^2\frac{x}{2}$ we have:
$$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\frac{2\sin^2\frac{x}{2}}{x^2}$$
$$=2\lim_{x\to 0}\frac{\sin\frac{x}{2}\cdot \sin\frac{x}{2}}{\frac{x}{2}\cdot\frac{x}{2}\cdot 4}=\frac{2}{4}=\frac{1}{2}$$
