Why is the limit of a function defined at only a cluster point of the domain?

Question

My textbook gives the following definition of a limit:

Let $A\subseteq\mathbb R$, and let $c$ be a cluster point of $A$. For a function $f:A\to\mathbb R$, a real number $L$ is said to be the limit of $f$ at $c$ if, given any $\varepsilon>0$, there exists $\delta>0$ such that if $x\in A$ and $0<|x-a|<\delta$, then $|f(x)-l|<\varepsilon$.

Why is it that definition of limit is always defined at the cluster point and not any other point of the domain?

Please answer in a way that is easy to understand because I have just started this course so I am not very familiar with topology.

Answer 1

Suppose that $e$ is not a cluster point. Then at an intuitive level, there is not a sequence of points "leading up to $e$", and for limits, we tend to think of points 'getting closer and closer to $e$', and then examine the values of our function $f$ at those points, and compare those values to $L$.

Think of a set like $X = [0, 1] \cup \{2\} \cup [3, 4]$. For any function $f$, no matter what value you picked for $L$, the definition of 'limit', applied at $x = 2$, would say that the limit of $f$ at $2$ was $L$. That's a little weird, because "the limit of $f$ at $2$" should be (at most!) a single value, not a whole range of them. To avoid this messiness, we rule out isolated points in the definition.

In particular, we can say "If the limit of $f$ at $x = c$ exists, it is unique." which is a very useful lemma to carry around in your pocket.

Answer 2

If you define limits whenever $a$ is in the domain of $f$, rather than whenever $a$ is a cluster point of the domain, then strange things can happen. Specificially, if $f:A\to\mathbb R$ is a function, and $a\in A$ is not a cluster point of $A$, then for every $l\in\mathbb R$, it is true that $f$ approaches $l$ near $a$. Why? If $a$ is not a cluster point, then there is some $d>0$ such that $(a-d,a+d)\cap A=\{a\}$. Now let $\varepsilon>0$ and set $\delta= d$. The statement

For all $x\in A$, if $0<|x-a|<\delta$ then $|f(x)-l|<\varepsilon$.

is vacuously true, since there is no $x\in A$ for which $0<|x-a|<\delta$ holds.

In addition, there are cases where limits should be defined despite the fact $a\not\in A$. A sensible definition of limits should assert that $\sin(x)/x\to 1$ as $x\to0$, but if we only define limits when $a$ is in the domain of $f$, then this is not true.

Answer 3

Given the definition above function f takes an element of A and results in an element of ℝ,

If the following conditions are true

• x is an element of A (x ∈ A),
• x is close (defined by ) to cluster point c (0 < |x - c| < )
• and f(x) is close (defined by ) to a real number L (|f(x) - L| < )

then real number L is said to be the limit of f at cluster point c.

For a limit to be applicable a value x is supposed to approach a value c, where c and x are in the same set.

Since x can approach c it means that c has to be a cluster point, x couldn’t approach c if c isn’t a cluster point. The cluster point c isn’t a specific point in the function, many points in the domain that can be approached with similar values are also cluster points.

Answer 4

Let $S$ be a subset of $\mathbb{R}$, and $f:S \rightarrow \mathbb{R}$. We say that the limit as $x$ approaches $c$ of $f(x)$ exists and is equal to $L$ if and only if for all $\epsilon >0$, there exists $\delta >0$ such that the following implication is true -

$\left[\ (x \in S) \ \land (0 < \lvert x-c \rvert < \delta) \ \right] \rightarrow \left(\ \lvert f(x) - L \rvert < \epsilon \ \right).$

Implicitly required in this $\epsilon-\delta$ definition is that $c$ is a cluster point of $f$. Normally, the issue is not explicit when $S=\mathbb{R}$, but ambiguities may arise if $S$ is a strict subset of $\mathbb{R}$. In fact, not requiring $c$ to be a cluster point allows us to pick the limit $L$ as anything we wish! The argument is as follows -

Suppose $c$ is not a cluster point of $S$. Then, there exists $\delta>0$ such that $(c-\delta,c+\delta) \ \bigcap \ S \ \backslash \ \{c\} = \varnothing$. Pick this choice of $\delta$. Then, for all $x \neq c$ in the interval $(c-\delta, c+\delta)$, it is easy to see that the antecedent in the above implication is false, hence we have a vacuous truth.

In summary, for a proper, well-defined singular limit to exist, it is necessary for the limiting value $c$ to be a cluster point of the domain $S$.