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

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

Proof
By the Definition of Limit of a Real Function, we have to prove that: $\forall \epsilon >0 \exists \delta >0 (\left|x-a \right|<\delta \implies \left|f\left(x\right)-L \right|<\epsilon)$

Therefore, let $\epsilon >0$ be given.

We have, also by the Definition of Limit of a Real Function:


 * (1) $\forall \epsilon' >0 \exists \delta >0 (\left|x-a \right|<\delta \implies \left|h\left(x\right)-L \right|<\epsilon)$


 * (2) $\forall \epsilon' >0 \exists \delta >0 (\left|x-a \right|<\delta \implies \left|g\left(x\right)-L \right|<\epsilon)$

And, since $\displaystyle \lim_{x \to a} \ g \left({x}\right)=\displaystyle \lim_{x \to a} \ h \left({x}\right)$, we have, by Combination Theorem for Limits of Functions/Sum Rule: $\displaystyle \lim_{x \to a} \ h \left({x}\right) - g \left({x}\right) =0$, and then:


 * (3) $\forall \epsilon' >0 \exists \delta >0 (\left|x-a \right|<\delta \implies \left|h\left(x\right)-g\left(x\right) \right|<\epsilon$

Take $\epsilon'=\frac{\epsilon}{3}$ in (1),(2),(3). Therefore, there exists $\delta_1, \delta_2, \delta_3$ that satisfies (1),(2),(3) with $\epsilon'=\frac{\epsilon}{3}$. Take $\delta=min\{\delta_1, \delta_2, \delta_3\}$

This way,


 * $\left|x-a\right|<\delta \implies \left|h\left(x\right)-L\right|<\frac{\epsilon}{3}, \left|g\left(x\right)-L\right|<\frac{\epsilon}{3}, \left|h\left(x\right)-g\left(x\right)\right|<\frac{\epsilon}{3}$

So, if $\left|x-a\right|<\delta$, we have that:


 * $\left|f\left(x\right)-L\right|=\left|f\left(x\right)-g\left(x\right)+h\left(x\right)-L+g\left(x\right)-h\left(x\right)\right|\leq \left|f\left(x\right)-g\left(x\right)\right|+\left|h\left(x\right)-L \right|+\left|h\left(x\right)-g\left(x\right)\right| \leq \left|h\left(x\right)-g\left(x\right)\right|+\left|h\left(x\right)-L \right|+\left|h\left(x\right)-g\left(x\right) \right|<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$