Find A and B in this limit Can you find $a$ and $b$? In how many ways can I find them?
$$\lim_{x\to0} \frac{a+\cos(bx)}{x^2}=-8$$ 

If the numerator is a number other than 0 the limit equals $\infty$. So in order to be equal to $-8$ the numerator has to be 0 which means that $\cos bx =-a  \iff bx=\pm \arccos(-a)+2k\pi \iff x=\pm \frac{\arccos(-a)+2k\pi}{b}$ where $k \in \mathbb{Z}$.This is the general form if you are interested(I wrote it just in case).
But here because $\cos bx=1$ , $\forall b\in \mathbb{R}$ it means that $a=-1$. Now in order to find $b$ you use the L'Hopital Rule.
Correct me if something is missing or wrong.
