Find $(a,b)$ such that $\lim\limits_{x\to 0}\frac{f(x)}{x}=1$ implies $\lim\limits_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$ 
Let $f$ denote any function such that $\lim\limits_{x\to 0}\frac{f(x)}{x}=1$. Find the value of $a$ and $b$, assuming that $\lim\limits_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$.

My  attempt:
$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$
$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3\frac{(f(x))^3}{x^3}}=1$
$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$
$\lim_{x\to 0}\frac{1+a\cos x-b\frac{\sin x}{x}}{x^2}=1$
$\lim_{x\to 0}\frac{1+a\cos x-b}{x^2}=1$
As denominator is tending to zero,so numerato will also tend to zero.
$1+a-b=0............................(1)$
Applying L Hospital rule,
$\lim_{x\to 0}\frac{-a\sin x}{2x}=1$
So $a=-2$ and $b=-1$
But the answer given in my book is $a=\frac{-5}{2}$ and $b=-\frac{3}{2}$.
I do not understand where have i gone wrong?

Since we can write 
$$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}\cdot\frac{1}{(f(x)/x)^3}$$
we have to have
$$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$$
We can write 
$$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}\tag1$$$$=\lim_{x\to 0}\frac{1+a\cos x-b\frac{\sin x}{x}}{x^2}\tag2$$
But we cannot write $(2)$ as
$$\lim_{x\to 0}\frac{1+a\cos x-b\cdot 1}{x^2}$$

From $(1)$, by L'Hôpital's rule,
$$(1)=\lim_{x\to 0}\frac{1+a\cos x-ax\sin x-b\cos x}{3x^2}\tag3$$
Here, we have to have
$$1+a-b=0\tag 4$$
Using L'Hôpital's rule several times,
$$\begin{align}(3)&=\lim_{x\to 0}\frac{1-ax\sin x-\cos x}{3x^2}\\&=\lim_{x\to 0}\frac{-a(\sin x+x\cos x)+\sin x}{6x}\\&=\lim_{x\to 0}\frac{-2a\cos x+ax\sin x+\cos x}{6}\\&=\frac{-2a+1}{6}\end{align}$$
and so
$$\frac{-2a+1}{6}=1\tag5$$
Now solve $(4)(5)$.
