Accumulation point in the definition of a limit of a function

Question

It is stated that the following is the definition of a limit of a function is:

Let f:D→ℝ be a function and let c be a accumulation point of D. Then we say L∈ℝ is the limit of f at c written lim_x→cf(x)=L if ∀ϵ>0 there exists a δ>0 such that ∀x∈D with 0<∣x−c∣<δ we have that ∣f(x)−L∣<ϵ.

I'm just wondering, how and why does the accumulation point play an important part in the definition?

Answer 1

Suppose that $c \not\in D'$. Since $c \not\in D'$, theres exist $\delta >0$ such that $V_{\delta} = (D\setminus \lbrace c \rbrace)\cap(c - \delta, c+ \delta) = \emptyset$. Thus, for all $\epsilon >0$, take this $\delta$. Therefore, $\emptyset = f(V_{\delta}) \subset (L - \epsilon, L+\epsilon)$ for all $L \in \mathbb{R}$, then, $\displaystyle \lim_{x\to c}f(x) = L$ for all $L \in \mathbb{R}$.

Note that if $c \not\in D'$, $\displaystyle \lim_{x\to c}f(x)$ would not be well defined.

$D'$ is the set of accumulation points.

Answer 2

If $c$ is not an accumulation point, then for some $\delta_0 > 0$, $B_{\delta_0}(c) = \{x: |x-c| < \delta_0\} = \{x\}$, in which case the notion of a limit existing at that point does not capture the essence of what one wants to define. In particular, the definition breaks down and becomes vacuously true and hence we see that $\lim_{x \to c} f(x) = L$ for every $L \in \mathbb{R}$.

Note that the definition of continuity of $f$ at an isolated point $c$ is such that $f$ is continuous at $c$ always (because $\{c\} = B_{\delta_0}(c)$ is open), regardless of how the function is defined on the rest of the metric space, which is consistent with the "limit" not meaning what we want it to mean, so we decide to not define it for isolated points.