<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] We need two limits below (which are easily obtained and the second one necessitates the use of Taylor series or L'Hospital's Rule) $$\lim_{x\to 0}\frac{1-\cos x} {x^{2}}=\frac{1}{2},\,\lim_{x\to 0}\frac{x-\sin x} {x^{3}}=\frac{1}{6}$$ The limit in question can be written as $$\lim_{x\to 0}(a+1)\cdot\frac{x-\sin x} {x^{3}}-a\frac{1-\cos x} {x^{2}}-\frac{(b-a-1)\sin x} {x^{3}}=1$$ This means that $$\lim_{x\to 0}\frac{(b-a-1)\sin x}{x^{3}}=1+\frac{a}{2}-\frac{a+1}{6}\tag{1}$$ or (upon multiplication by $x^{2}$) $$\lim_{x\to 0}\frac{(b-a-1)\sin x} {x} =0$$ ie $b-a-1=0$. Using this in $(1)$ we get $$1+\frac{a}{2}-\frac{a+1}{6}=0$$ ie $a=-5/2,b=a+1=-3/2,a+b=-4$.

Your approach has a fundamental issue when you replace $(b\sin x) /x^{3}$ by $b/x^{2}$. This is not allowed by any theorem on limits and is an invalid step in the evaluation of a limit. 
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