Limit of Tan X over X at Zero/Proof 3

Proof
Let $f$ be the real function defined as:
 * $\map f x = \sin x$

Let:
 * $c = \pi$
 * $h \in \openint 1 {\dfrac \pi 2}$

We have:

Hence:

Hence by the Squeeze Theorem:


 * $\ds \lim_{h \mathop \to 0} \dfrac {\tan h} h = 1$

The result follows on renaming $h$ to $x$.