Squeeze Theorem/Functions/Proof 3

Theorem
Let $a$ be a point on an open real interval $I$.

Also let $f$, $g$ and $h$ be real functions defined at all points of $I$ except for possibly at point $a$.

Suppose that:
 * $\forall x \ne a \in I: g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$
 * $\displaystyle \lim_{x \mathop \to a} \ g \left({x}\right) = \lim_{x \mathop \to a} \ h \left({x}\right) = L$

Then:
 * $\displaystyle \lim_{x \mathop \to a} \ f \left({x}\right) = L$

Proof
By the definition of the limit of a real function, we have to prove that:
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{f \left({x}\right) - L}\right| < \epsilon}\right)$

Let $\epsilon \in \R_{>0}$ be given.

We have:
 * $\displaystyle \lim_{x \mathop \to a} \ g \left({x}\right) = \lim_{x \mathop \to a} \ h \left({x}\right)$

Hence by Sum Rule for Limits of Functions:
 * $\displaystyle \lim_{x \mathop \to a} \ h \left({x}\right) - g \left({x}\right) = 0$

By the definition of the limit of a real function:


 * $(1): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{h \left({x}\right) - L}\right| < \epsilon'}\right)$
 * $(2): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{g \left({x}\right) - L}\right| < \epsilon'}\right)$
 * $(3): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{h \left({x}\right) - g \left({x}\right)}\right| < \epsilon'}\right)$

Take $\epsilon' = \dfrac {\epsilon} 3$ in $(1)$, $(2)$, $(3)$.

Then there exists $\delta_1, \delta_2, \delta_3$ that satisfies $(1)$, $(2)$, $(3)$ with $\epsilon' = \dfrac \epsilon 3$.

Take $\delta = \min\{\delta_1, \delta_2, \delta_3\}$.

Then:

So, if $\left\vert{x - a}\right\vert<\delta$: