If \(a\) and \(b\) are non-negative real numbers and
\[
\lim_{x\to0}\frac{e^{ax}-\cos bx}{1-\cos x}=4,
\]
then
\[
\lim_{x\to a}\frac{\sin(bx-ab)}{x-a}
=
\]
Show Hint
Remember the standard limits
\[
\boxed{
\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12,
\qquad
\lim_{x\to0}\frac{\sin x}{x}=1.
}
\]
Also, if a limit is finite, first eliminate any lower-order terms in the numerator.