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?
 A: 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?
A: Rewrite $$A=\frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}= \frac{x(1+a\cos(x))-b\sin(x)}{x^3}\times\Big(\frac{x}{f(x)}\Big)^3$$ and now, as Olivier Oloa answered, use Taylor expansions.
Whichever could be $\lim_{x\to0} \frac{f(x)}{x}=L$ (except $0$ or $\infty$), you could find the conditions you want for  $\lim_{x\to0} A=M$.
A: 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 $a, b$.
A: Since $\displaystyle\lim_{x\to0}\frac{f(x)}{x}=1$, we have
$$\lim_{x\to0}\frac{x}{f(x)}
=\lim_{x\to0}\frac{1}{\frac{f(x)}{x}}
=\frac{1}{\displaystyle\lim_{x\to0}\frac{f(x)}{x}}
=\frac{1}{1}=1.$$
Now, by using L'Hopital's rule, we have
\begin{align}
1&=\lim_{x \to 0} \frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}\\
&=\lim_{x \to 0}\left[\frac{x^3}{f(x)^3}\cdot\frac{x(1+a\cos(x))-b\sin(x)}{x^3}\right]\\
&=\left(\lim_{x \to 0}\frac{x}{f(x)}\right)^3\cdot
\lim_{x\to0}\frac{x(1+a\cos(x))-b\sin(x)}{x^3}\\
&\stackrel{\rm H}{=}
1^3\cdot\lim_{x\to0}\frac{1+a\cos(x)-ax\sin(x)-b\cos(x)}{3x^2}\tag{1}\\
&\stackrel{\rm H}{=}
\lim_{x\to0}\frac{-a\sin(x)-a\sin(x)-ax\cos(x)+b\sin(x)}{6x}\\
&\stackrel{\rm H}{=}
\lim_{x\to0}\frac{-2a\cos(x)-a\cos(x)+ax\sin(x)+b\cos(x)}{6}\\
&=\frac{-3a+b}{6}\tag{2}.
\end{align}
By $(1)$, we have to force the limit of the numerator to be zero, so $1+a-b=0$. Also, by $(2)$ we naturally get $-3a+b=6$. Hence we conclude that
$a=-\frac{5}{2}$ and $b=-\frac{3}{2}$.
