the derivative of $\sin{x}$ is $\cos{x}$ I can  prove the  $\sin'{x}=\cos{x}$ by formula $\lim_{h\to 0}{\frac{f(x+h)-f(x)}{h}}$ but the proof is not known for me by the formula 
 $\lim_{z\to x}{\frac{\sin{z}-\sin{x}}{z-x}}$?
Can anyone give me the hint of its proof?
 A: HINT:
Using Prosthaphaeresis Formula, 
$$\sin z-\sin x=2\sin\frac{z-x}2\cos\frac{z+x}2 $$
Use $\lim_{h\to0}\dfrac{\sin h}h=1$
A: Here is a straightforward way to think
$${\sin(z) - \sin(x)\over z - x} = {\sin(x + (z - x)) - \sin(x)
\over z - x}. $$
Do you see it now?
A: by computing the perimeter of a regular polygon lying between its inscribed and exscribed circles we have:
$$
\frac{\pi}n\cos\frac{\pi}n \lt \sin \frac{\pi}n\lt \frac{\pi}n
$$
which, with continuity, assures us that $\lim_{h\to0}\dfrac{\sin h}h=1$
now look at
$$
\lim_{h\to0}\dfrac{\sin (x+h)-\sin (x-h) }{2h} = \cos x  \lim_{h\to 0} \frac{\sin h}{h}$$
the compound angle formulae employed here are easily deduced from the definitions using elementary geometry
A: I propose this proof based on the differential calculus. We start from the definition
$$\sin\left(x\right):=\sum_{k\in\mathbb{N}}\left(-1\right)^{k}\frac{x^{2k+1}}{\left(2k+1\right)!}.$$
Let $x\in\mathbb{R}$. We choose $h\in\mathcal{B}_{\left|.\right|}\left(0,\eta\right)$
 for $\eta<1$. We have :
$$\sin\left(x+h\right)=\sum_{n=0}^{+\infty}\left(-1\right)^{n}\sum_{k=0}^{2n+1}\frac{x^{2n+1-k}h^{k}}{k!\left(2n+1-k\right)!}$$
 $$=\sum_{n=0}^{+\infty}\left(-1\right)^{n}\frac{x^{2n+1}}{\left(2n+1\right)!}+\sum_{n=0}^{+\infty}\left(-1\right)^{n}\frac{x^{2n}}{\left(2n\right)!}h+\sum_{n=1}^{+\infty}\left(-1\right)^{n}\sum_{k=2}^{2n+1}\frac{x^{2n+1-k}h^{k}}{k!\left(2n+1-k\right)!}$$
$$=\sin\left(x\right)+h\cos\left(x\right)+\mathcal{R}_{x}\left(h\right).$$
The map $h\mapsto h\cos\left(x\right)$ is clearly linear and continuous from $\mathcal{B}_{\left|.\right|}\left(0,\eta\right)$
  to $\mathbb{R}$. As for the rest, we have :
$$\left|\mathcal{R}_{x}\left(h\right)\right|\leq\sum_{n=1}^{+\infty}\sum_{k=2}^{2n+1}\frac{\left|x\right|^{2n+1-k}\left|h\right|^{k}}{k!\left(2n+1-k\right)!}=\left|h\right|^{2}\sum_{n=1}^{+\infty}\sum_{k=2}^{2n+1}\frac{\left|x\right|^{2n+1-k}\left|h\right|^{k-2}}{k!\left(2n+1-k\right)!}$$
$$=\left|h\right|^{2}\sum_{n=1}^{+\infty}\sum_{p=0}^{2n-1}\frac{\left|x\right|^{2n-1-p}\left|h\right|^{p}}{\left(p+2\right)!\left(2n-1-p\right)!}\leq\left|h\right|^{2}\sum_{n=1}^{+\infty}\sum_{p=0}^{2n-1}\frac{\left|x\right|^{2n-1-p}\left|h\right|^{p}}{p!\left(2n-1-p\right)!}$$
$$=\left|h\right|^{2}\sum_{n=1}^{+\infty}\frac{\left(\left|x\right|+\left|h\right|\right)^{2n-1}}{\left(2n-1\right)!}
 =\left|h\right|^{2}\sum_{m=0}^{+\infty}\frac{\left(\left|x\right|+\left|h\right|\right)^{2m+1}}{\left(2m+1\right)!}$$
$$\leq\left|h\right|^{2}\sum_{m=0}^{+\infty}\frac{\left(\left|x\right|+\left|h\right|\right)^{m}}{m!}=\left|h\right|^{2}\exp\left(\left|x\right|+\left|h\right|\right)=\mathcal{O}\left(\left|h\right|^{2}\right).$$
Hence, when $\left|h\right|\rightarrow0$
 we have $\left|\mathcal{R}_{x}\left(h\right)\right|\rightarrow0$
  and thus $\mathrm{d}\sin\left(x\right)=\cos\left(x\right)$.
