Solving limit given a limit Given the fact that 
$$\lim_{h\to 0} \frac{\sin(h)}{h}=1,$$
compute the following limit: 
$$\lim_{h\to 0}\frac{\sin(x+h) - \sin(x)}{h}\ $$
How would I go about solving this problem? I have attempted to use trig identities (addition of angles) to try to simplify the problem, but it only seemed to make it worse...
 A: \begin{align*}
\lim_{h \rightarrow 0} \frac{\sin(x+h)-\sin x}h &=\lim_{h \rightarrow 0}  \frac{(\sin x \cos h + \cos x \sin h)-\sin x}h & \text{trigonometric sum formula} \\
&=\lim_{h \rightarrow 0}  \frac{\sin x(\cos h-1) + \cos x \sin h}h &\text{shuffle terms in numerator} \\
&=\lim_{h \rightarrow 0}  \left(\frac{\sin x(\cos h-1)}h + \frac{\cos x \sin h}h \right) & \text{break the fraction} \\
&=\lim_{h \rightarrow 0}  \frac{\sin x(\cos h-1)}h +\lim_{h \rightarrow 0}  \frac{\cos x \sin h}h \\
&=\sin x \lim_{h \rightarrow 0}  \frac{\cos h-1}h + \cos x \lim_{h \rightarrow 0}  \frac{\sin h}h \\
&= \sin x \cdot 0 + \cos x \cdot 1 & \text{apply limit indentities}\\
&= \cos x.  & \text{simplify}
\end{align*}
A: $$
\sin(x+h)=\sin x\cos h+\sin h\cos x,
$$
and hence
$$
\frac{\sin(x+h)-\sin x}{h}=\frac{\sin x\cos h+\sin h\cos x-\sin x}{h}=\cos x+\sin x\frac{\cos h -1}{h} \\= \cos x-\sin x\frac{2\sin^2(h/2)}{h}=\cos x-\sin x \cdot \frac{h}{2} \cdot\left(\frac{\sin(h/2)}{h/2}\right)^2 \to \cos x-\sin x\cdot 0 \cdot 1\\=\cos x
$$
A: Use $\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)$
A: Using the difference to product identity $\sin A - \sin B = 2\sin \dfrac{A-B}{2}\cos\dfrac{A+B}{2}$, we get: 
$\displaystyle\lim_{h \to 0}\dfrac{\sin(x+h)-\sin x}{h} = \lim_{h \to 0}\dfrac{2\sin\frac{h}{2}\cos(x+\frac{h}{2})}{h} = \lim_{h \to 0}\dfrac{\sin \frac{h}{2}}{\frac{h}{2}} \cdot \lim_{h \to 0}\cos(x+\tfrac{h}{2}) = $...
A: This is the limit definition of the derivative for $\sin x$
So,
$$(\sin x)' = \cos x$$
EDIT, given any $f(x)$, to find its derivative, through limits we can express it as:
$$\large \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = f'(x)$$
Here $f(x) = \sin x$
