$\lim_{x \to 0}\frac{x- \sin{x}}{x^2}$ Calculate:
$$\lim_{x \to 0}\frac{x- \sin{x}}{x^2}$$
I would like to try but i don't find any idea i don't know how to use Hopital rule i tried to return $\cos$ to $\sin$
But it doesn't work a help
 A: you can prove that the limit exist bu several way
so to find the limit without l'hopital :
$$l=\lim_{x\to 0}\frac{x-\sin x}{x^2}\stackrel{x\to3x}{=}\frac{1}{9}\lim_{x\to 0}\frac{3x-3\sin x+4\sin^3x}{x^2}$$
$$l=\frac{1}{3}\lim_{x\to 0}\frac{x-\sin x}{x^2}+\frac{4}{9}\lim_{x\to 0}\frac{\sin^3x}{x^2}$$
$$l=\frac{1}{3}l+0\Rightarrow l=0$$
A: The "right" thing to use is the Taylor series, but we will work our way around that. We also avoid L'Hospital's Rule. By the Mean Value Theorem, for any $x\ne 0$ there is a $c_x$ between $0$ and $x$ such that 
$$\frac{x-\sin x}{x}=1-\cos c_x=2\sin^2(c_x/2).$$
Thus
$$\frac{\sin x-x}{x^2}=\frac{2\sin^2(c_x/2)}{x}.$$
Suppose $|x|$ is small non-zero. Since $|c_x/2|\lt |x|/2$ we have
$$0\lt \frac{2\sin^2(c_x/2)}{|x|}\lt \frac{x^2/2}{|x|}\lt |x|/2.$$
The right-hand side approaches $0$ as $x\to 0$, so our limit is $0$.
A: L'Hopital is the easy way to do this, but here's another approach, using only $(\sin x)'=\cos x$ and $\sin x\lt x$ (for $x$ positive):
Since ${x-\sin x\over x^2}$ is an odd function, it suffices to consider the one-sided limit with $x\gt0$.  We have
$$0\le{x-\sin x\over x^2}={1\over x^2}\int_0^x(1-\cos u)du={1\over x^2}\int_0^x{\sin^2u\over1+\cos u}du\le{1\over x^2}\int_0^x\sin^2udu\\={1\over x^2}\int_0^xu^2{\sin^2u\over u^2}du\le{1\over x^2}\int_0^xu^2du={1\over x^2}\cdot{1\over3}x^3={1\over3}x\to0\text{ as }x\to0^+$$
A: You can use l'Hopital's rule if you know how to take derivatives.  After checking that the fraction is naively $0/0$, differentiate the top of the fraction, and the bottom of the fraction.
$$
\frac{1-\cos x}{2x}$$
Still $0/0$?  Then do it again:
$$
\frac{\sin x}{2} \to 0$$
SO the answer will be zero.  Are you sure the problem did not read $$\lim_{x \to 0}\frac{x- \sin{x}}{x^3}$$, which requires three applicatoins and gives an answer of $\frac16$?
