Proof of limits involving the epsilon/delta with 0 ≤ x ≤ 8, ≤ confusion

Question

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.

Answer 1

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 $\delta = \text{min}\left(1,\dfrac{\epsilon}{9}\right)$

Answer 2

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.