Given $\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$, what is the value of $a+b$? 
Given that
  $$\lim\limits_{x\to 0} \frac {x(1+a\cos x)-b\sin x}{x^3}=1$$
  What is the value of $a+b$

My try 
$\lim\limits_{x\to 0} (\frac {x(1+a\cos x)}{x^3}-\frac {b\sin x}{x^3})=1$
$\lim\limits_{x\to 0} (\frac {(1+a×cos(x)}{x^2}-\frac {b}{x^2})=1$
$\lim\limits_{x\to 0} \frac {1+a\cos x-bx}{x^2}=1$
Apply L'Hôpital's rule: 
$\lim\limits_{x\to 0} \frac {-a\cos x-b}{2x}=1$
Apply L'Hôpital's rule again: 
$\lim\limits_{x\to 0} \frac {-a\sin x}{2}=1$ $\to$  $a=-2$
Is my approach right? 

It is:
$$L=\lim_{x\to 0} \frac {x+ax\cos x-b\sin x}{x^3}\stackrel{LR}=\lim_{x\to 0} \frac {1+a\cos x-ax\sin x-b\cos x}{3x^2}\stackrel{1+a-b=0; LR}=\lim_{x\to 0} \frac {-a\sin x-a\sin x-ax\cos x+b\sin x}{6x}\stackrel{LR}=\lim_{x\to 0} \frac {-a\cos x-a\cos x-a\cos x+ax\sin x+b\cos x}{6}=1 \Rightarrow -3a+b=6.$$
Hence:
$$\begin{cases}1+a-b=0 \\ -3a+b=6\end{cases} \Rightarrow a+b=-\frac52-\frac32=-4.$$
