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? 

As $x(1+a\cos x)-b\sin x\sim (1+a-b)x$ as $x\to0$ then the given limit
can only exist when $b=1+a$. A series expansion then gives
$$x(1+a\cos x)-(1+a)\sin x\sim-\frac{a}2x^3+(1+a)\frac{x^3}6.$$
One needs then $(1+a)/6-a/2=1$ for the given limit to equal $1$.
Now one can obtain $a$ and $b$.
