<s>[INST] 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?
 [/INST] Just as mathlove answered, we need to find $a,b$ such that $$\lim_{x\to 0}\frac{x(1+a\cos (x))-b\sin (x)}{x^3}=1$$ Let us use the classical Taylor expansions $$\cos(x)=1-\frac{x^2}{2}+O\left(x^3\right)$$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ and replace to get $$\frac{x(1+a\cos (x))-b\sin (x)}{x^3}=\frac{x (a-b+1)+x^3 \left(\frac{b}{6}-\frac{a}{2}\right)+O\left(x^4\right)}{x^3}$$ So, the first thing is $a-b+1=0$ and the second (for a limit equal to $1$) is $\frac{b}{6}-\frac{a}{2}=1$. Solving these two equations leads to the required values.
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