You are making it entirely too difficult. First of all, you are asked to find $F'(0)$ so $x=0$ and you might better start with$$
F'(0) = \lim_{h\to 0} \frac{F(0+h) - F(0)}{h} = \lim_{h\to 0} \frac {f(h)\sin^2(h)}{h}
$$where I have omitted terms that are $0$. Now think about breaking that into the product of three factors, each of which is easy to find the limit.
Thanks for you help! After mulling over your explanation I understand how you arrived at:
When you say break it into a product of three factors I assume you simply mean:
If this is correct I still wouldn't know how to continue from here, as in all the examples iv'e come across before the h in the denominator cancels. Currently we would be dividing by zero if we took the limit of the function. My only other thought was to rewrite sin^2 (x) in some other fashion using trig identities but that still doesn't help with the h in the denominator.
Thanks a bunch again.
No, that is NOT what he meant! For one thing you have mysteriously changed two of the "h"s to "x"s. Also you did not completely change it into a product of three terms.
You should have $\displaystyle \left(\lim_{h\to 0} f(h)\right)\left(\lim_{h\to 0} sin(h)\right)\left(\lim_{h\to 0}\frac{sin(h)}{h}\right)$.