Evaluate the limit of $(f(2+h)-f(2))/h$ as $h$ approaches $0$ for $f(x) = \sin(x)$. If $f(x)=\sin x$, evaluate $\displaystyle\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$ to two decimal places. 
Have tried to answer in many different ways, but always end up getting confused along the way, and am unable to cancel terms. 
Any help is always appreciated! 

Use Prosthaphaeresis Formulas,
$$\sin(a+h)-\sin a=2\sin\dfrac h2\cos\dfrac{2a+h}2$$
$$\lim_{h\to0}\dfrac{\sin(a+h)-\sin a}h=\lim_{h\to0}\dfrac{\sin\dfrac h2}{\dfrac h2}\cdot\lim_{h\to0}\cos\dfrac{2a+h}2=?$$
