f(x) is a function such that $\lim_{x\to0} f(x)/x=1$ $f(x)$ is a function such that $$\lim_{x\to0} \frac{f(x)}{x}=1$$ if 
$$\lim_{x \to 0} \frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}=1$$
Find $a$ and $b$
Can I assume $f(x)$ to be $\sin(x)$ since $\sin$ satisfies the given condition?

Hint. You may use the standard Taylor series expansions, as $x \to 0$,
$$
\begin{align}
\cos x&=1-\frac{x^2}2+O(x^4)\\
\sin x&=x-\frac{x^3}6+O(x^4)
\end{align}
$$ giving
$$
\begin{align}
\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}&=\frac{(1+a-b) x+\frac16 (-3 a+b) x^3+O(x^5)}{(f(x))^3}
\\\\&=\frac{(1+a-b) x+\frac169 (-3 a+b) x^3+O(x^5)}{x^3(1+\epsilon(x))^3}
\end{align}
$$ where, as $x \to 0$, we have used $f(x)=x(1+\epsilon(x))$ with $\epsilon(x) \to 0$.
Can you take it from here?
