Boundedness of Sine X over X

Theorem
Let $$x \in \mathbb{R}$$.

Then $$\left|{\frac {\sin x} {x}}\right| \le 1$$.

Proof
From Derivative of Sine Function, we have $$D_x \left({\sin x}\right) = \cos x$$.

So by the Mean Value Theorem, there exists $$\xi \in \mathbb{R}$$ between $$0$$ and $$x$$ such that:

$$\frac {\sin x - \sin 0} {x - 0} = \cos \xi$$.

From Boundedness of Sine and Cosine we have that $$\left|{\cos \xi}\right| \le 1$$.