Find A and B in this limit Can you find $a$ and $b$? In how many ways can I find them?
$$\lim_{x\to0} \frac{a+\cos(bx)}{x^2}=-8$$ 
 A: If you apply L'Hospital rule twice, you'll end up with: 
$$\lim_{x -> 0} \frac{b^2\cos(bx)}{2} = 8,$$ 
and using various properties of limits you'll be able to arrange it to solve for b, which I'll leave up to you. As for a, you'll most likely spot it quite quickly.
But to prepare you for the future: L'Hospital rule can be used how ever many times you'ld like. Depending on context​ of course, but you get it.
A: Yes, I can find $a$ and $b$ ;-) A hint :
$$\cos(bx)=1-\frac{(bx)^2}{2}+o(x^2).$$
A: 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
$a = -1, b=4$.
A: 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.
