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! 

Remember that $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
Here we see that $f(x) = \sin x$ and $x = 2$. Therefore, all we have to do is find $f'(2)$. If you know that the derivative of the sine function is $\cos x$ then you immediately have the answer as $\cos (2)$ (Remember that this is in Radians... all of Calculus should be done in radians)
