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? 
 A: 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
$$\begin{cases}b-3a=6\\1+a-b=0\end{cases}\implies a=-\frac52,b=-\frac32\\\implies a+b=-4$$
A: 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:
$$\begin{cases}1+a-b=0 \\ -3a+b=6\end{cases} \Rightarrow a+b=-\frac52-\frac32=-4.$$
A: 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 $b$.
A: 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. 
