Squeeze Theorem/Functions/Proof 3

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

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

We have:
 * $\ds \lim_{x \mathop \to a} \map g x = \lim_{x \mathop \to a} \map h x$

Hence by Sum Rule for Limits of Real Functions:
 * $\ds \lim_{x \mathop \to a} \map h x - \map g x = 0$

By the definition of the limit of a real function:


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

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 $\size {x - a} < \delta$: