Why does the definition of the functional limit involve a limit point?

Question

This may be an odd question, and I'm not sure if these type of questions are at all appreciated in the maths community. But given the definition of the functional limit:

Let $f: A \to \mathbb{R}$, and let $c$ be a limit point of the domain $A$. We say that $\lim_{x \to c} f(x)=L$ provided that, for all $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $0<|x-c|< \delta$ (and $x \in A$) it follows that $|f(x)-L|< \epsilon$.

why does $c$ have to be a limit point? I understand that the above is just a definition, and we can define anything we want to. But, after having gone through an entire chapter of functional limits and continuity, I have not yet figured out what would go wrong if $c$ was not a limit point.

Answer 1

One important property of limits, is that if a limit exists, then its value is unique.

Specifically, there should be at most one number $L$ such that $\forall \epsilon > 0, \exists \delta > 0, \forall x \in A, 0 < |x-c| < \delta \implies |f(x)-L| < \epsilon$.

Now, lets suppose $c$ is not a limit point of $A$. Then, there exists some radius $r$ such that the interval $(c-r,c+r)$ contains no points in $A$ other than $c$ itself.

Then, for any function $f$, pick your favorite number $L$. Let $\epsilon > 0$ be arbitrary and choose $\delta = r$. Then, since there are no points $x \in A$ such that $0 < |x-c| < \delta = r$, the statement $\forall x \in A, 0 < |x-c| < \delta \implies |f(x)-L| < \epsilon$ is "vacuously true".

Therefore, $\displaystyle\lim_{x \to c}f(x) = L$ holds for any number $L$ that you picked, i.e. $\displaystyle\lim_{x \to c}f(x)$ can be any number.

This is why we insist on $c$ being a limit point of $A$.

Answer 2

For the definition to be reasonable, we want that for every $\delta>0$ there is $x\in A$ so that $0<|x-c|<\delta$. This condition means exactly that $c$ is a limit point of $A$.

Answer 3

If $c$ is not a limit point of $A$, then for all small enough $\delta > 0$, the set $\{ x \in A : 0 < \lvert x-c\rvert < \delta\}$ is empty, and the condition

$$(x\in A \land 0 < \lvert x-c\rvert < \delta) \implies \lvert f(x) - L\rvert < \varepsilon$$

is vacuously satisfied for every $L$. Thus the limit would not be unique if we didn't require $c$ to be a limit point of $A$.

Answer 4

To see where things go wrong take $A=\mathbb{N}$ and some function $f:A\to\mathbb{R}$ and say you want to prove that $$ \lim_{x\to c}f(x)=L $$

where $c\in A$ is a natural number.

Let $\epsilon>0$ and choose $\delta(\epsilon)=\delta=\frac{1}{2}$

Indeed - For every $x\in A$ s.t $0<|x-c|<\frac{1}{2}$ we have it that $|f(x)-L|<\epsilon$

This is true vacuously because there are no such $x\in A$ that satisfy $0<|x-c|<\frac{1}{2}$ so we get $$ \lim_{x\to c}f(x)=L $$

regardless of what $L$ is, which is unreasonable