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! 

"[I] am unable to cancel terms."
This suggests you're trying to do some algebra.  But the phrase "to two decimal places" suggests that doing arithmetic was intended instead.  The means you must somehow find $\sin 2$ and $\sin2.01$ or some other value of the sine for a number near $2$.  If we assume calculators are infallible, we get
$$
\frac{\sin(2+0.01) - \sin 2}{0.01} \approx \frac{0.9050906 - 0.9092974}{0.01} \approx -0.4206864.
$$
Since $\cos2\approx-0.4161468$, this is within $0.01$ of the right value.
Remember: This is in radians, not in degrees.
This limit is usually found not by doing cancelations, but by squeezing.
