Boundedness of Sine X over X

Theorem

Let $x \in \R$.

Then:

$\left|{\dfrac {\sin x} x}\right| \le 1$

Proof

From Derivative of Sine Function, we have:

$D_x \left({\sin x}\right) = \cos x$

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

$\dfrac {\sin x - \sin 0} {x - 0} = \cos \xi$

From Boundedness of Sine and Cosine we have that:

$\left|{\cos \xi}\right| \le 1$

$\blacksquare$