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

For
$\lim_{x\to 0}\frac{\ a+\cos bx}{x^2}=-8
$,
note that,
for small $x$,
$\cos x
\approx 1-\frac{x^2}{2}
$.
Therefore
$\dfrac{\ a+\cos bx}{x^2}
\approx
\dfrac{\ a+1-\frac{(bx)^2}{2}}{x^2}
$.
If
$a+1 \ne 0$,
then
$\dfrac{\ a+1-\frac{(bx)^2}{2}}{x^2}
\to \infty$
as $x \to 0$.
Therefore,
to have the limit exist,
we must have
$a+1=0$
or $a = -1$.
The expression then becomes
$\dfrac{\ a+1-\frac{(bx)^2}{2}}{x^2}
=\dfrac{-\frac{(bx)^2}{2}}{x^2}
=-\frac{b^2}{2}
$.
If this is $-8$,
then
$b^2 = 16$,
so 
$b = 4$.
Therefore
$a = -1, b=4$.
