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: we have:  $\dfrac{1+a\cos x}{x^2} - \dfrac{b}{\sin^2 x} \to 1$, and rewrite $\dfrac{b}{\sin^2 x} = \dfrac{b}{x^2}\cdot \dfrac{x^2}{\sin^2 x}$ then the limit on the left equals $\dfrac{1+a\cos x - b}{x^2} = 1$, this means you can assume L'hopitale rule meaning: $1 + a - b = 0$, and differentiate both numerator and denominator to get another equation and solve for $a, b$.
