Question

lim x->0 [cos(ax)-b]/(2x^2) = -1, solve for a and b

Answer 1

@Kainui should we assume we can use l'hopital rule?

Answer 2

\[\lim_{x\to0}\frac{\cos ax-b}{2x^2}=-1\]
The only way you would be able to do this is if L'hopital's was allowed, otherwise the limit would not exist due to the zero denominator.

In order to get an appropriate indeterminate form, you must have \(\cos ax-b\to0\) as \(x\to0\):
\[\lim_{x\to0}(\cos ax-b)=\cos 0-b=\cdots\]
Solving for \(b\) is easy from here.

Now apply L'Hopital's to the limit:
\[\lim_{x\to0}\frac{\cos ax-1}{2x^2}=\lim_{x\to0}\frac{-a\sin ax}{4x}=\lim_{x\to0}\frac{-a^2\cos ax}{4}=-\frac{a^2}{4}=-1\]
Solving for \(a\) is also easy.