Hi, can someone help me with this?

lim x->0 $\displaystyle \frac{x-tan(4x)}{sin(4x)}$

I tried multiplying by the conjugate then using the trig identity for tan, but I got a bunch of mess I don't know what to do with.

Also, for

lim x->0 $\displaystyle \frac{x^2-12x+32}{sin(x-4)}$

is the limit -8? just want to make sure that's right.

and how do I solve this:

let F(x) = h(g(f(x))), where f(4) = 7, g(7) = 8, f '(4) = 3, g '(7) = 8, and h '(8) = 2

then find F'(4)

confused as how to do this

thanks!