Why is $|x - c| < \delta$ used in epsilon-delta proofs?

Question

$\varepsilon$-$\delta$ definition of a limit:

For every $\varepsilon$ $>$ $0$, there exists $\delta$ $>$ $0$ such that whenever $0$ $<$ $|x-c|$ $<$ $\delta$, we have $|f(x)-L|$ $<$ $\varepsilon$.

My doubt arised from the following source:

Prove, using delta and epsilon, that $\lim_{x \to 5} ( 3x^{2} - 1 ) = 74$

Feel free to get the bigger picture: refer to "Examples using a Non-Linear Function", highlighted in blue. Note that the picture above it's only part of the proof, which is defining $\varepsilon$

How To Construct a Delta-Epsilon Proof. (n.d.). Milefoot. Retrieved April 27, 2021, from http://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm

According to what I'ved learned, we must prove that $0 < |x - c| < \delta$ $\implies$ $0 \leq |f(x) - L| < \varepsilon $, not $|x - c| < \delta$ $\implies$ $0 \leq|f(x) - L| < \varepsilon $

Answer 1

The given solution makes the assumption that $f$ is continuous at $c$, meaning $L = \lim_{x \to c} f(x) = f(c)$. With this assumption, for $x = c$ we still have $\lvert f(x) - L\rvert < \epsilon$ because $L = f(c)$.

If $f$ is not continuous at $c$, meaning $L \ne f(c)$, then we could have $f(x)$ is close to $f(c)$ for all $x$ close to but not equal $c$, but at exactly $f(c)$ we have a discontinuity.

Here's a picture from Brilliant.org I've modified (where $x_0 = c$) to include a discontinuity at $f(x_0)$, making $f(x_0)$ totally different from its neighboring values and showing the $\epsilon-\delta$ definition clearly doesn't hold.

Answer 2

Consider this function: $$f(x) = \begin{cases} x+2, & x \ne 1 \\ 0, & x = 1, \end{cases}$$ and suppose we are interested in $$L = \lim_{x \to 1} f(x).$$ It is easy to see that we should have $L = 3$. The epsilon-delta definition says that for any $\epsilon > 0$, there exists $\delta > 0$ such that whenever $0 < |x - 1| < \delta$, then $|f(x) - L| < \epsilon$. And we can see in this case that we do in fact require $0 < |x-1|$, since if not, we could choose $\epsilon = 10^{-100}$, and $x = 1$, but $|f(x) - L| = |0 - 1| = 1 > \epsilon$.

The purpose of $\delta$ is to establish a neighborhood around the limit point $c$ such that whenever $x$ is within a ball of radius $d$ but is not equal to $c$, then the value of the function for all points inside that ball will not deviate from $L$ by more than some chosen $\epsilon$. So in order to exclude the limit point itself, we can require $0 < |x - c| < \delta$, or we can write something to the effect of "for any $\epsilon > 0$, there exists $\delta > 0$ such that whenever $x \ne c$ and $|x - c| < \delta$, then..." But the effect is the same.

Answer 3

Definition:

Let $A\subseteq\mathbb{R}$ be non-empty and suppose $f:A\to\mathbb{R}$ is a function. For $a,L\in\mathbb{R}$, we say that $\lim\limits_{x\to a}f(x)=L$ if and only if $$\forall\varepsilon>0\exists\delta>0\forall x\in A:0<|x-a|<\delta\Longrightarrow |f(x)-L|<\varepsilon$$ Note that $a$ need not be in $A$ and/or $f$ need not be defined at $a$.

Examples:

1. Say $A=(-1,0)\cup (0,1)$ and $f:A\to\mathbb{R}$ is $f(x)=1$ for all $x\in A$. Then $\lim\limits_{x\to 0}f(x)=1$, but $0\not\in A$.

2. Say $A=\mathbb{R}$ and $f:A\to\mathbb{R}$ is $f(x)=1$ for $x\neq 0$ and $f(x)=0$ for $x=0$. Then $\lim\limits_{x\to 0}f(x)=1$.

Definition:

Let $A\subseteq\mathbb{R}$ be non-empty and suppose $f:A\to\mathbb{R}$ is a function. For $a\in A$, we say that $f$ is continuous at $a$ if and only if $\lim\limits_{x\to a}f(x)=f(a)$, i.e, $$\forall\varepsilon>0\exists\delta>0\forall x\in A:0<|x-a|<\delta\Longrightarrow |f(x)-f(a)|<\varepsilon$$

Note that Example 2. shows that the limit of a function can exist at a point, but the function need not be continuous, or even defined at that point.

In the definition of continuity at a point, the second definition, we must have $a\in A$ for this definition to make sense, otherwise $|f(x)-f(a)|$ is meaningless, as in Example 1. In the case where $|x-a|=0$ i.e., $x=a$, everything is fine, since the function $f$ must be defined on all of $A$, so $|f(x)-f(a)|=|f(a)-f(a)|=0$, so we can loosen our definition of continuity to allow for $|x-a|=0$, since the function is defined at that point.

The answer by heropup and the comments under qwr's answer show examples of what happen if you alter the definition of a limit to exsit at a point.