Limits with L'Hôpital's rule Find the values of $a$ and $b$ if $$ \lim_{x\to0} \dfrac{x(1+a \cos(x))-b \sin(x)}{x^3} = 1 $$
I think i should use L'Hôpital's rule but it did not work.

The easiest way (in my opinion) is to plug in the power series expansion of $x(1+a\cos x)-b\sin x$ around zero. Then, the limit becomes
$$\lim_{x\rightarrow 0} \frac{x(a-b+1)+x^3(b-3a)/6+O(x^5)}{x^3}=1.$$
Now you have two equations involving $a$ and $b$ to satisfy (can you figure out what those equations are?)
