Proof of limits involving the epsilon/delta with 0 ≤ x ≤ 8, ≤ confusion I need help with a proof similar to the following: 

Consider the function $f(x) = x^2$ for $0 \leq x \leq 8$. Prove that $ \lim_{x \to 4} ⁡f(x)=16$.

I understand how to do it for $0 < x < 8$, but the $=$ part of the signs are confusing me. Particularly as we usually start: Proof: Let $\epsilon > 0$ be given. For $\delta > 0$ such…. Since a new delta forces $f(x)$ to be in the new epsilon can $= 0$ ever be true? I cannot find any examples anywhere that show an example working with ≤ and I don't know how to handle this.
 A: I think the proof goes in the same way for $0<x<8$ and $0\leq x \leq 8$ as others commented and asnwered. 
Let's say $D$ is some domain around $x=4$. For example, $D$ may be $(0,8)$, $[0,8]$, or $\mathbb{R}$. Let's consider 
\begin{align}
f(x) =& x^2& \text{ for } x \in& D.
\tag{1}
\label{eq: 1}
\end{align} 
The statement,
\begin{equation}
\lim_{x\rightarrow 4} f(x) = 16,
\tag{2}
\label{eq: 2}
\end{equation}
means that
for any positive number $\epsilon$, there exists another positive number $\delta$ such that 
\begin{align}
x\in& D& \text{ and }&&
|x-4| <& \delta&  \Rightarrow&&
|f(x) - 16| <& \epsilon.
\tag{3}
\label{eq: 3}
\end{align}
Let's prove this.
The condition $|x-4| < 0$ is equivalent to 
\begin{equation}
4 -\delta < x < 4 + \delta.
\tag{4}
\label{eq: 4}
\end{equation}
If $0<4-\delta$, i.e., if $\delta<4$, the above expression implies that
\begin{equation}
16 -8\delta +\delta^2 < x^2 < 16 + 8\delta + \delta^2,
\tag{5}
\label{eq: 5}
\end{equation}
which is equivalent to 
\begin{equation}
-8\delta +\delta^2 < x^2-16 < 8\delta + \delta^2.
\tag{6}
\label{eq: 6}
\end{equation}
This implies
\begin{equation}
|x^2-16| < 8\delta + \delta^2
\tag{7}
\label{eq: 7}
\end{equation}
since $|-8\delta +\delta^2| < |8\delta + \delta^2|$ for $\delta>0$.
If $\delta<1$, then $\delta^2 < \delta$ and $8\delta + \delta^2< 9\delta$.
Therefore,
\begin{equation}
|x^2-16| < 9\delta.
\tag{8}
\label{eq: 8}
\end{equation}
So far we have assumed that $\delta < 4$ and $\delta < 1$. We are free to impose these restrictions on $\delta$ because we are requested to show the existence of $\delta$ that satisfy a given condition and we are making the condition more strict by ourselves.
From these restrictions and eq. (\ref{eq: 8}), we see that eq. (\ref{eq: 3}) is satisfied if we take $\delta$ as
\begin{equation}
\delta = \min \left\{\frac{\epsilon}{9}, 1\right\}.
\tag{9}
\label{eq: 9}
\end{equation}
This completes the proof. 
By saying 'any positive number $\epsilon$', we are mainly interested in an arbitrarily small value of $\epsilon$, but just in case the value of $\epsilon$ is given $9$ or larger, we can always return $\delta=1$ in order to satisfy eq. (\ref{eq: 3}).
For given $D$ and $\epsilon$, if
\begin{equation}
D \subset
\left\{x\big||x-4|<\min\left\{\frac{\epsilon}{9},1\right\}\right\},
\end{equation}
then $|f(x)-16| < \epsilon$ for all $x \in D$, and whatever $\delta>0$ satisfy eq. (\ref{eq: 3}). So, setting $\delta$ as in eq. (\ref{eq: 9}) is certainly fine.
The discussion above does not change as long as $D$ includes a neighbourhood of $x=4$. So, as long as $f(x)$ is defined in the neighbourhood of $x=4$, the limit, $\lim_{x\rightarrow 4} f(x)$, does not change.
A: Case 1: $ 0 \le x \le 8$ :$|f(x)-16| = |x^2-16| = |x-4||x+4| \le 12|x-4| < \epsilon \iff |x-4| < \dfrac{\epsilon}{12}$. Thus take $\delta = \dfrac{\epsilon}{12}$. 
Case 2: $ x \in \mathbb{R}$: the difference is to choose $\delta < 1$, then $|x+4| \le |x-4|+ |8| < 1+8 = 9 \implies $ choose $\delta = \text{min}\left(1,\dfrac{\epsilon}{9}\right)$
