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...
 A: 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.
A: $$\sin x-x=\int_0^x(\cos t-1)\,dt=\int_0^x(-2\sin^2(t/2))\,dt$$
Using $\vert\sin\theta\vert\le\vert\theta\vert$, we have:
$$\vert \sin x-x\vert \le \left\vert \int_0^x 2(t/2)^2\,dt\right\vert = \frac{\vert x \vert^3}{6}$$
A: The strategy depends on the choices offered. 
Let us look at $|x|\le 1$. The Maclaurin series for $\sin x$ goes like this:
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots.$$
For $|x|\le 1$ this is an alternating series, so the truncation error is less in absolute value than the first neglected term. Thus
$$|\sin x-x|\le \frac{|x^3|}{6}$$
if $|x|\le 1$. So $C=\frac{1}{6}$ works for $|x|\le 1$. From your calculation, $C=2$ works for $x\gt 1$, so $C=2$ works for all $x$.
A: In general if we truncate the Taylor series (centered at zero) for an analytic function $f$ at the $k$th term, then the remainder can be expressed in the form
$$R(x) = \frac{f^{k+1}(t)}{(k+1)!}x^{k+1}$$
where $t$ is some number which depends on $x$. Let's apply this to $f(x) = \sin(x)$ with $k=2$. Then the truncated Taylor series is
$$\sum_{k=0}^{2}\frac{f^{k}(0)}{k!}x^k = x$$
(the $k=0$ and $k=2$ terms are zero). Then the remainder is
$$\begin{aligned}
R(x) &= \sin(x) - \sum_{k=0}^{2}\frac{f^{k}(0)}{k!}x^k\\
& = \sin(x) - x\\
\end{aligned}$$
which can be expressed using the formula above, with $k=2$:
$$R(x) = \frac{f^{3}(t)}{3!}x^{3} = \frac{-\cos(t)}{3!}x^3$$
As $|\cos(t)| \leq 1$, we obtain the bound
$$|\sin(x) - x| = |R(x)| \leq \frac{1}{3!}|x^3|$$
as desired.
A: Take the power series of sine : $$\forall x \in \mathbb{R},\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5}-\frac{x^7}{7!}+\dots$$
$$|\sin(x)-x|=|-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots|\le|x^3||-\frac{1}{3!}+\frac{x^2}{5!}-\frac{x^4}{7!}+\dots|$$
By the ratio test $(-\frac{1}{3!}+\frac{x^2}{5!}-\frac{x^4}{7!}+\dots)$ converges and its limit is : $-\frac{2-2 e^{-x^2}-2 x^2+x^4}{2 x^6}$ (you notice that this series is similar to the power series of $e^{-x^2}$ and you complete to obtain exactly the series wanted) which is bounded by $C=1$ for example. Finally : $|\sin(x)-x|\le C|x^3|$.
