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?
 A: You need to use the Taylor expansion of $\sin(x)$. The result pretty much follows straight from there.
A: Start with
$\sin' = \cos
$,
$\cos' = -\sin
$,
$\sin(0) = 0$,
$\cos(0) = 1$,
and
$\sin^2+\cos^2 = 1$.
For small $t$,
$1 \ge \cos(t)
\ge 0
$
so
$\sin(x)
=\int_0^x \cos(t)dt
\le x
$.
Therefore
$1-\cos(x)
=\int_0^x \sin(t) dt
\le \int_0^x t dt 
= \frac{x^2}{2}
$
so
$\cos(x)
\ge 1-\frac{x^2}{2}
$.
Therefore
$\sin(x)
=\int_0^x \cos(t)dt
\ge\int_0^x (1-\frac{t^2}{2})dt
=x-\frac{x^3}{6}
$.
So we already have
$x-\frac{x^3}{6}
\le \sin(x)
\le x
$.
This is actually
enough for what you want.
Doing this again,
$1-\cos(x)
=\int_0^x \sin(t) dt
\ge \int_0^x (t-\frac{t^3}{6}) dt 
= \frac{x^2}{2}-\frac{x^4}{24}
$
so
$\cos(x)
\ge 1-\frac{x^2}{2}+\frac{x^4}{24}
$.
Doing this one more time,
$\sin(x)
=\int_0^x \cos(t)dt
\ge\int_0^x (1-\frac{t^2}{2}+\frac{t^4}{24})dt
=x-\frac{x^3}{6}+\frac{x^5}{120}
$.
By induction,
we can get the power series
for $\sin$ and $\cos$.
Note:
This is not original.
I first saw this in
"100 Great Problems of Elementary Mathematics"
by Heinrich Dorrie 
(less than $15 from Dover).
Get it.
A: 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 $$\frac{\sin h}{h} = 1 + O(h^2) .$$
A: If you know that $\cos(x)=1+O(x^2)$ and $|\tan x|>|x|$ then you know that $\sin x = \tan x \cos x = x+O(x^3)$.
