<s>[INST] 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? 
 [/INST] Series methods are always useful.
$$\cos(x)=1-\frac12x^2+\mathcal O(x^3)\\1+a\cos(x)=1+a-\frac a2x^2+\mathcal O(x^3)\\x(1+a\cos(x))=(1+a)x-\frac a2x^3+\mathcal O(x^4)\\\sin(x)=x-\frac16x^3+\mathcal O(x^4)\\x(1+a\cos(x))-b\sin(x)=(1+a-b)x+\frac{b-3a}6x^3+\mathcal O(x^4)\\\frac{x(1+a\cos(x))-b\sin(x)}{x^3}=\frac{1+a-b}{x^2}+\frac{b-3a}6+\mathcal O(x)$$
For this to go to $1$ as $x\to0$, we need $1+a-b=0$ and $\frac{b-3a}6=1$, which gives us
$$\begin{cases}b-3a=6\\1+a-b=0\end{cases}\implies a=-\frac52,b=-\frac32\\\implies a+b=-4$$
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