How to prove $\sin(x) = x + O(x^3)$? I was doing an exercise in which I have to find the rate of convergence of 
$$\lim\limits_{h\to 0}\dfrac{\sin h}{h} = 1$$
and the answer is $O(h^2)$. I don't understand why. The only thing I have found is that $$\sin(x) = x + O(x^3)$$ when $x$ tends to zero, and with that the exercise is solved, but how I demonstrate that fact?

One can see this quickly using the given limit, that $\sin$ is (1) an odd function (which follows more or less immediately from the usual exponential and unit circle definitions) and is (2) twice differentiable:
The limit and the differentiability imply that
$$\sin h = h + a h^2 + O(h^3),$$
and oddness implies that $a = 0$. So, $\sin h = h + O(h^3)$ and thus $$\frac{\sin h}{h} = 1 + O(h^2) .$$
