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.

You can, unless using L'Hospital's rule repeatedly, which  is not the alpha and omega of limit computations. The simplest way to go is to use Taylor's polynomial:
$$ \cos x=1-\dfrac{x^2}2+o(x^2),\quad \sin x=x-\frac{x^3}6+o(x^3),$$
whence 
$$x(1+a\cos x)-b\sin x=(1+a-b)x+\Bigl(\frac b6-\frac a2\Bigr)x^3+o(x^3) $$
$$\frac{x(1+a\cos x)-b\sin x}{x^3}= \frac{(1+a-b)x+\Bigl(\dfrac b6-\dfrac a2\Bigr)x^3+o(x^3)}{x^3}. $$
For the limit to be equal to $1$, the following equations must be satisfied:
$$\begin{cases}
a-b+1=0,\\\dfrac b6-\dfrac a2=1.
\end{cases}$$
