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

$\blacksquare$