Limits and Continuity

Donovan D Chang

20 Limits and Continuity

Again, we revisit limits and continuity. However, this time, we define the limit with a term, accumulation point, and focus on the limit of real-valued functions. Therefore, let us first define the accumulation point.

Definition. (Accumulation Point)
For a set $S\subseteq \mathbb{R}$ , a point $x\in \mathbb{R}$ is an accumulation point of $S$ if every deleted neighborhood of $x$ contains a point of $S$ .
The deleted neighborhood of $x$ refers to the neighborhood exclusive of the point of origin itself, i.e.

$\mathcal{N}^*(x,\delta) = \mathcal{N}(x,\delta) \setminus \left\lbrace x\right\rbrace$

Alternative terms for accumulation point are limit point or cluster point. Also, the accumulation point need not be in $S$ .

Definition. (Limit)
For an accumulation point $c$ of a set $D$ , given a mapping $f:D\rightarrow\mathbb{R}$ , a real number $L$ is a limit of $c$ , if for every $\epsilon>0$ there exists $\delta >0$ such that

$0<|x-c|<\delta \Rightarrow |f(x)-L|<\epsilon$

Note that the notion of deleted neighborhood is inherent in $0<|x-c|<\delta$ , i.e. $x\neq c$ .

Formal definition of continuity at $x=c$ , provided in the calculus section, is obtained by requiring $f(c)=L$ and allowing for the possibility $x=c$ in the previous definition of a limit. Here we introduce another intuitive way of expressing the continuity using the notion of neighborhood.

Theorem.
The statement $f$ of a mapping $f:D\rightarrow\mathbb{R}$ is continuous at $c$ is equivalent to
for every neighborhood $V$ of $f(c)$ , there exists a neighborhood $U$ of $c$ such that $f(U\cap D)\subseteq V$ .

Note the equivalence of the two definitions [12]

One of most important properties of a continuous function is the intermediate value theorem.

Theorem. (Intermediate Value Theorem)
Given a continuous function $f:[a,b]\rightarrow \mathbb{R}$ , where $f(a)\neq f(b)$ ,
the intermediate value theorem warrants the existence of $c\in (a,b)$ such that $f(c)=k$ , where $\text{min}\left(f(a),f(b)\right)<k<\text{max}\left(f(a),f(b)\right)$

Let us illustrate with a straightforward example.

Example. Determine whether there is a root of $\sin{x}=0$ on $\left[ -\frac{\pi}{2}, \frac{\pi}{2}\right]$ .

Let $f(x)=\sin{x}$ .
Then, $f\left(-\frac{\pi}{2}\right)=-1$ and $f\left(\frac{\pi}{2}\right)=1$ .
In fact, $0\in (-1,1)$ .
Therefore, by the intermediate value theorem, there exists at least one root of $\sin{x}=0$ on $\left[ -\frac{\pi}{2}, \frac{\pi}{2}\right]$ .

Uniform Continuity

We conclude this section with uniform continuity, a different kind of continuity from an ordinary continuity. Formally, uniform continuity is defined as

Definition. (Uniform Continuity)
Given a function $f:D\rightarrow \mathbb{R}$ , the function $f$ is uniformly continuous on $D$ if for every $\epsilon>0$ there exists $\delta >0$ such that

$0<|x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon$

Roughly speaking, uniform continuity guarantees the proximity of the two values of the function, $f(x)$ and $f(y)$ , given the proximity of $x,y\in D$ , whereas the maximum distance between $f(x)$ and $f(y)$ depends on $x,y\in D$ themselves in ordinary continuous functions.

Let us illustrate with an example.

Example. Examine whether $f(x)=x^2$ is uniformly continuous.

$\left\vert f(x)-f(y)\right\vert = \left\vert x^2-y^2\right\vert = |x+y||x-y|$

Let $\epsilon = 1$ .

If we let $y=x+\frac{\delta}{2}$ , then $|x-y|=\frac{\delta}{2}<\delta$ .
Therefore, by letting $|x+y|\geq \frac{2}{\delta}$ , we can make

$\left\vert f(x)-f(y)\right\vert = \frac{\delta}{2}\cdot |x+y| \geq 1$

Then, if we let $x=\frac{1}{\delta}$ and $y=\frac{1}{\delta}+\frac{\delta}{2}$ ,

$\begin{align*} \left\vert f(x)-f(y)\right\vert &= \left\vert x^2-y^2\right\vert\\ &= |x+y||x-y|\\ &= \left\vert \frac{1}{\delta}+\frac{1}{\delta}+\frac{\delta}{2}\right\vert \cdot \frac{\delta}{2}\\ &> \frac{2}{\delta} \cdot \frac{\delta}{2} = 1 \end{align*}$

Therefore, $f$ is not uniformly continuous on $\mathbb{R}$ .

20 Limits and Continuity

Uniform Continuity

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