There exists a constant $C$ such that $|\sin(x)-x|\le C|x^3|$ for all real $x$ Problem:

Show that there exists a constant $C$ such that $|\sin(x)-x|\le C|x^3|$ for all real $x$.

I'm stuck on this question. First, I noticed that $|\sin(x)-x|\le 1+|x|$. But I'm not sure how to deal with $|x|\le 1$.
Let $f(x)=|\sin(x)-x|$ and $g(x)=|x^3|$. I noticed that $f(0)=g(0)$, so I looked into derivatives. For a moment I only looked at when $0<x<1$. Then since $f(x)<0$, $f'(x)=1-\cos(x)$. On the other hand, $g'(x)=3x^2$. Then I'm stuck. 
This comes from a multiple choice question of a GRE Math Subject practice problem so it shouldn't take very long...

Let:
$$f(x) = \begin{cases} \frac{\sin x - x}{x^3}, x \neq 0 \\ -1/6, x = 0 \end{cases}$$
$f$ is continuous on $\Bbb R$ and:
$$\lim_{x \to \pm \infty} f(x) = 0$$
Which shows that $f$ is bounded.
