Proving continuity – "Math for Non-Geeks"

Overview

There are several methods for proving continuity:

• Concatenation Theorems: If the function can be written as a concatenation of continuous functions, it's continuous by the Concatenation Theorems.
• Using the local nature of continuity: If a function looks like another well-known continuous function in a small neighborhood of a point, then it must also be continuous in this point.
• Considering the left- and right-sided limits: If one can show that the left- and right-sided limits of a function are the same in some point, then the function is continuous in this point.
• Showing the sequence criterion: Using the sequence criterion means that the limit can be pulled into the function, i.e. we can consider the limit of the arguments. For a sequence $(x_n{)}_{n\in \mathbb{N}}$ of arguments with limit value $x_0$ it must hold $\underset{n\to \mathrm{\infty }}{lim}f(x_n)=f(x_0)=f\left(\underset{n\to \mathrm{\infty }}{lim}{x}_n\right)$.
• Showing the epsilon-delta criterion: For every $ϵ>0$ it must be show that there exists some $\delta >0$ such that for all arguments $x$ with a distance smaller than $\delta$ from the point $x_0$, the inequality $|f(x)-f(x_0)|<ϵ$ is satisfied.

Composition of continuous functions

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General proof sketch

Following the concatenation theorems, every composition of continuous functions is again continuous function. So if $f:D\to ℝ$ can be written as a concatenation of continuous functions, we can directly infer continuity of $f$ . A corresponding proof could be of the following form:

Let $f:\dots$ with $f(x)=\dots$. The function $f$ is a concatenation of the following functions:

...List of continuous functions, which serve as building bricks for $f$ ...

Since $f(x)=\dots$ (Expression how $f$ is constructed out of those bricks) , we know that $f$ is a concatenation of continuous functions and hence continuous, as well.

In a lecture, this proof scheme may of course only be applied if the concatenation theorems have already been treated before. In any way, it is a very efficient method to characterize continuous functions.

Example Problem

Exercise (Epsilon-Delta proof for the continuity of the Square Root Function)

Show, using the epsilon-delta criterion, that the following function is continuous:

$f:ℝ\to ℝ,x\mapsto \sqrt{5+x^2}$

How to get to the proof? (Epsilon-Delta proof for the continuity of the Square Root Function)

We need to show, that for any given $ϵ>0$ , there is a $\delta >0$ , such that all $x\in ℝ$ with $|x-a|<\delta$ satisfy the inequality $|f(x)-f(a)|<ϵ$ . So let us take a look at the target inequality $|f(x)-f(a)|<ϵ$ and estimate the absolute $|f(x)-f(a)|$ from above. We are able to control the term $|x-a|$ . Therefor, would like to get an upper bound for $|f(x)-f(a)|$ including the expression $|x-a|$ . So we are looking for an inequality of the form

$|f(x)-f(a)|\le K(x,a)\cdot |x-a|$

Here, $K(x,a)$ is some expression depending on $x$ and $a$ . The second factor is smaller than $\delta$ and can be made arbitrarily small by a suitable choice of $\delta$ . Such a bound is constructed as follows:

$\begin{array}{cc}|f(x)-f(a)|& =|\sqrt{5+x^2}-\sqrt{5+a^2}|\\ & \downarrow \text{expand with}|\sqrt{5+x^2}+\sqrt{5+a^2}|\\ & =\frac{|\sqrt{5+x^2}-\sqrt{5+a^2}|\cdot |\sqrt{5+x^2}+\sqrt{5+a^2}|}{|\sqrt{5+x^2}+\sqrt{5+a^2}|}\\ & \downarrow \sqrt{5+x^2}+\sqrt{5+a^2}\ge 0\\ & =\frac{|x^2-a^2|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & =\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\cdot |x-a|\\ & \downarrow K(x,a):=\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & =K(x,a)\cdot |x-a|\end{array}$

Since $|x-a|<\delta$ , there is:

$|f(x)-f(a)|=K(x,a)\cdot |x-a|<K(x,a)\cdot \delta$

If we now choose $\delta$ small enough, such that $K(x,a)\cdot \delta \le ϵ$ , then we obtain our target inequality $|f(x)-f(a)|\le ϵ$. But $K(x,a)$ still depends on $x$ , so $\delta$ would have to depend on , too - and we required one choice of $\delta$ which is suitable for all $x$ . Therefore, we need to get rid of the $x$-dependence. This is done by an estimate of the first factor, such that our inequality takes the form $K(x,a)\le \tilde{K}(a)$ :

$\begin{array}{cc}K(x,a)& =\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & \downarrow \text{triangle inequality}\\ & \le \frac{\left|x\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}}+\frac{\left|a\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & \downarrow \frac{a}{b+c}\le \frac{a}{b}\text{ for }a,c\ge 0,b>0\\ & \le \frac{\left|x\right|}{\sqrt{5+x^2}}+\frac{\left|a\right|}{\sqrt{5+a^2}}\\ & \downarrow \frac{|a|}{\sqrt{5+a^2}}\le 1\text{, since }\sqrt{5+a^2}\ge \sqrt{a^2}=|a|\\ & \le 2=:\tilde{K}(a)\end{array}$

We even made $\tilde{K}(a)$ independent of $a$ , which would in fact not have been necessary. So we obtain the following inequality

$|f(x)-f(a)|\le 2\cdot |x-a|<2\cdot \delta$

We need the estimate $2\cdot \delta \le ϵ$, in order to fulfill the target inequality $|f(x)-f(a)|<ϵ$ . The choice of $\delta =\frac{ϵ}{2}$ is sufficient for that. So let us write down the proof:

Proof (Epsilon-Delta proof for the continuity of the Square Root Function)

Let $f:ℝ\to ℝ$ with $f(x)=\sqrt{5+x^2}$. Let $a\in ℝ$ and an arbitrary $ϵ>0$ be given. We choose $\delta =\frac{ϵ}{2}$. For all $x\in ℝ$ with $|x-a|<\delta$ there is:

$\begin{array}{cc}|f(x)-f(a)|& =|\sqrt{5+x^2}-\sqrt{5+a^2}|\\ & =\frac{|\sqrt{5+x^2}-\sqrt{5+a^2}|\cdot |\sqrt{5+x^2}+\sqrt{5+a^2}|}{|\sqrt{5+x^2}+\sqrt{5+a^2}|}\\ & =\frac{|x^2-a^2|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & =\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\cdot |x-a|\\ & \le (\frac{\left|x\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}}+\frac{\left|a\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}})\cdot |x-a|\\ & \le (\frac{\left|x\right|}{\sqrt{5+x^2}}+\frac{\left|a\right|}{\sqrt{5+a^2}})\cdot |x-a|\\ & \le (1+1)\cdot |x-a|\\ & \le 2\cdot |x-a|\\ & \downarrow |x-a|<\delta =\frac{ϵ}{2}\\ & <2\cdot \frac{ϵ}{2}=ϵ\end{array}$

Hence, $f$ is a continuous function.

Using the Local Nature of Continuity

If applicable, one can use the fact that continuity is a local property. Namely, when a function $f$ looks the same as another function $g$ in a certain point $x$, i.e. $f(x)=g(x)$ on some small neighborhood of $x$, and we know $g$ to be continuous, then $f$ must also be continuous at $x$. For example, consider the function $f:ℝ\setminus \{0\}\to ℝ$ with $f(x)=1$ for positive $x$ and $f(x)=-1$ for negative $x$. Now we choose some random positive number $x_0$. In a sufficiently small neighborhood of $x_0$, $f$ is constant $1$:

The function f is constantly 1 in a sufficiently small neighborhood of a positive argument.

Since constant functions are continuous, $f$ is also continuous at the point $x_0$. Similarly, one can show that $f$ is also locally constant for negative number. Then $f$ is also continuous for negative numbers, and is therefore continuous for all real numbers. In the proof we write:

Such an argument can often be applied to functions that are defined by a case differential. Our function $f$ is a good example of that. In conclusion, it's defined as:

$f:ℝ\setminus \{0\}\to ℝ:x\mapsto \{\begin{array}{cc}1& x>0\\ 0& x<0\end{array}$

However, the local nature of continuity can not be used as an argument for every function defined by a case differential! Let's consider the following function $g$:

$g:ℝ\to ℝ:x\mapsto \{\begin{array}{cc}{x}^2& x\ge 0\\ |x|& x<0\end{array}$

For all points that are not zero we can construct a proof like we've done earlier in this chapter to show that the function is continuous at these points. However, at the point $x_0=0$ we have to construct a different argument. For example, here we could consider the left- and right-sided limits.

Sequence Criterion

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Review: Sequence Criterion

Definition (Sequence criterion of continuity)

A function $f:D\to ℝ$ with $D\subseteq ℝ$ is continuous, if for all $x\in D$ and all sequences $(x_n{)}_{n\in \mathbb{N}}$ with $\mathrm{\forall }n\in \mathbb{N}:x_n\in D$ and $\underset{n\to \mathrm{\infty }}{lim}{x}_n=x$ there is:

$\underset{n\to \mathrm{\infty }}{lim}f(x_n)=f\left(\underset{n\to \mathrm{\infty }}{lim}{x}_n\right)=f(x)$

General Proof Structure

In order to prove continuity of a function at some $x_0$ , we need to show that for each sequence $(x_n{)}_{n\in \mathbb{N}}$ of arguments converging to $\underset{n\to \mathrm{\infty }}{lim}{x}_n=x_0$ , there is $\underset{n\to \mathrm{\infty }}{lim}f(x_n)=f(x_0)$ . A proof for this could schematically look as follows:

Let $f:\dots$ be a function defined by $f(x)=\dots$ and $x_0=\dots$ . In addition, let $(x_n{)}_{n\in \mathbb{N}}$ be any sequence of arguments satisfying $\underset{n\to \mathrm{\infty }}{lim}{x}_n=x_0$. Then, there is:

$\underset{n\to \mathrm{\infty }}{lim}f(x_n)=\dots =f(x_0)$

In order to prove continuity for the function $f$ (for all arguments in its domain of definition), we need to slightly adjust that scheme:

Let $f:\dots$ be a function defined by $f(x)=\dots$ and let $x_0$ be any element of the domain of definition for $f$. In addition, let $(x_n{)}_{n\in \mathbb{N}}$ be any sequence of arguments satisfying $\underset{n\to \mathrm{\infty }}{lim}{x}_n=x_0$. Then, there is:

$\underset{n\to \mathrm{\infty }}{lim}f(x_n)=\dots =f(x_0)$

Example Problem

Exercise (Continuity of quadratic functions)

Show that the quadratic function $f:ℝ\to ℝ:x\mapsto x^2$ is continuous.

Proof (Continuity of quadratic functions)

Let $f:ℝ\to ℝ:x\mapsto x^2$. Consider any sequence $(x_n{)}_{n\in \mathbb{N}}$, converging to $x$ . There is

$\begin{array}{cc}\underset{n\to \mathrm{\infty }}{lim}f(x_n)& =\underset{n\to \mathrm{\infty }}{lim}{x}_n^2\\ & =\underset{n\to \mathrm{\infty }}{lim}{x}_n\cdot x_n\\ & \downarrow \underset{n\to \mathrm{\infty }}{lim}{a}_n\cdot b_n=\underset{n\to \mathrm{\infty }}{lim}{a}_n\cdot \underset{n\to \mathrm{\infty }}{lim}{b}_n\\ & =\left(\underset{n\to \mathrm{\infty }}{lim}{x}_n\right)\cdot \left(\underset{n\to \mathrm{\infty }}{lim}{x}_n\right)\\ & \downarrow \text{assumption}\underset{n\to \mathrm{\infty }}{lim}{x}_n=x\\ & =x\cdot x=x^2=f(x)\end{array}$

So we may pull the limit inside the brackets for the quadratic function and hence, it is continuous.

Epsilon-Delta Criterion

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Review: Epsilon-Delta Criterion

Definition (Epsilon-Delta-definition of continuity)

A function $f:D\to ℝ$ with $D\subseteq ℝ$ is continuous at $x_0\in D$, if and only if for any $ϵ>0$ there is a $\delta >0$ , such that $|f(x)-f(x_0)|<ϵ$ holds for all $x\in D$ with $|x-x_0|<\delta$ . Written in mathematical symbols, that means $f$ is continuous at $x_0\in D$ if and only if

$\mathrm{\forall }ϵ>0\mathrm{\exists }\delta >0\mathrm{\forall }x\in D:|x-x_0|<\delta ⟹|f(x)-f(x_0)|<ϵ$

General Proof Structure

In mathematical quantifiers, the epsilon-delta definition of continuity of a function $f$ at the point $x_0$ reads:

$\mathrm{\forall }ϵ>0\mathrm{\exists }\delta >0\mathrm{\forall }x\in D:|x-x_0|<\delta ⟹|f(x)-f(x_0)|<ϵ$

This technical method of writing the claim specifies the general proof structure for proving continuity using the epsilon-delta criterion:

$\begin{array}{c}\underset{\mathrm{\forall }ϵ>0}{\underbrace{\text{Let }ϵ>0\text{ be arbitrary.}}}\underset{\mathrm{\exists }\delta >0}{\underbrace{\text{Choose }\delta =\dots \text{ There is }\delta \text{, since}\dots }}\\ \underset{\mathrm{\forall }x\in D:|x-x_0|<\delta }{\underbrace{\text{Let }x\in D\text{ with }|x-x_0|<\delta \text{ be arbitrary.}}}\underset{⟹|f(x)-f(x_0)|<ϵ}{\underbrace{\text{There is: }|f(x)-f(x_0)|<\dots <ϵ}}\end{array}$

Example Problem and General Procedure

Exercise (Continuity of the quadratic function)

Prove that the function $f:ℝ\to ℝ$ with $f(x)=x^2$ is continuous.

How to get to the proof? (Continuity of the quadratic function)

For this proof, we need to show that the square function is continuous at any argument $x_0\in ℝ$ . Using the proof structure for the epsilon-delta criterion, we are given an arbitrary $ϵ>0$ . Our job is to find a suitable $\delta >0$ , such that the inequality $|f(x)-f(x_0)|<ϵ$ holds for all $|x-x_0|<\delta$.

In order to find a suitable $\delta$ , we plug in the definition of the function $f(x)=x^2$ into the expression $|f(x)-f(x_0)|$ which shall be smaller than $ϵ$:

$|f(x)-f(x_0)|=|x^2-x_0^2|$

The expression $|x-x_0|$ may easily be controlled by $\delta$. Hence, it makes sense to construct an upper estimate for $|x^2-x_0^2|$ which includes $|x-x_0|$ and a constant. The factor $|x-x_0|$ appears if we perform a factorization using the third binomial formula:

$|x^2-x_0^2|=|x+x_0||x-x_0|$

The requirement $|x-x_0|<\delta$ allows for an upper estimate of our expression:

$|x+x_0||x-x_0|<|x+x_0|\cdot \delta$

The $\delta$ we are looking for may only depend on $ϵ$ and $x_0$ . So the dependence on $x$ in the factor $|x+x_0|\cdot \delta$ is still a problem. We resolve it by making a further upper estimate for the factor $|x-x_0|$ . We will use a simple, but widely applied "trick" for that: A $x_0$ is subtracted and then added again at another place (so we are effectively adding a 0) , such that the expression $x-x_0$ appears:

$|x+x_0|\cdot \delta =|x\underset{=0}{\underbrace{-x_0+x_0}}+x_0|\cdot \delta =|x-x_0+2x_0|\cdot \delta$

The absolute $|x-x_0|$ is obtained using the triangle inequality. This absolute $|x-x_0|$ is again bounded from above by $\delta$ :

$|x-x_0+2x_0|\cdot \delta \le (|x-x_0|+|2x_0|)\cdot \delta <(\delta +2|x_0|)\cdot \delta$

So reshaping expressions and applying estimates, we obtain:

$|f(x)-f(x_0)|<(\delta +2|x_0|)\cdot \delta$

With this inequality in hand, we are almost done. If $\delta$ is chosen in a way that $(\delta +2|x_0|)\cdot \delta \le ϵ$ , we will get the final inequality $|f(x)-f(x_0)|<ϵ$ . This $\delta$ is practically found solving the quadratic equation ${\delta }^2+2|x_0|\delta =ϵ$ for $\delta$ . Or even simpler, we may estimate $(\delta +2|x_0|)\cdot \delta$ from above. We use that we may freely impose any condition on $\delta$ . If we, for instance, set $\delta \le 1$, then $\delta +2|x_0|\le 1+2|x_0|$ which simplifies things:

$|f(x)-f(x_0)|<(\underset{\le 1+2|x_0|}{\underbrace{\delta +2|x_0|}})\cdot \delta \le (1+2|x_0|)\cdot \delta$

So $(1+2|x_0|)\cdot \delta \le ϵ$ will also do the job. This inequality can be solved for $\delta$ to get the second condition on $\delta$(the first one was $\delta \le 1$):

$(1+2|x_0|)\cdot \delta \le ϵ⟺\delta \le \frac{ϵ}{1+2|x_0|}$

So any $\delta$ fulfilling both conditions does the job: $\delta \le 1$ and $\delta =\frac{ϵ}{1+2|x_0|}$ have to hold. Ind indeed, both are true for $\delta :=min\{1,\frac{ϵ}{1+2|x_0|}\}$. This choice will be included into the final proof:

Proof (Continuity of the quadratic function)

Let $ϵ>0$ be arbitrary and $\delta :=min\{1,\frac{ϵ}{1+2|x_0|}\}$. If an argument $x$ fulfills $|x-x_0|<\delta$ then:

$\begin{array}{cc}|f(x)-f(x_0)|& =|x^2-x_0^2|\\ & =|x+x_0||x-x_0|\\ & \downarrow |x-x_0|<\delta \\ & <|x+x_0|\cdot \delta \\ & =|x\underset{=0}{\underbrace{-x_0+x_0}}+x_0|\cdot \delta \\ & =|x-x_0+2x_0|\cdot \delta \\ & \le (|x-x_0|+2|x_0|)\cdot \delta \\ & \downarrow |x-x_0|<\delta \\ & <(\delta +2|x_0|)\cdot \delta \\ & \downarrow \delta \le 1\\ & \le (1+2|x_0|)\cdot \delta \\ & \downarrow \delta =\frac{ϵ}{1+2|x_0|}\\ & =(1+2|x_0|)\cdot \frac{ϵ}{1+2|x_0|}\\ & =ϵ\end{array}$

This shows that the square function is continuous by the epsilon-delta criterion.

Concatenated absolute function

Exercise (Example for a proof of continuity)

Prove that the following function is continuous at $x_0=1$:

$f:ℝ\to ℝ:x\mapsto f(x)=5|x^2-2|+3$

How to get to the proof? (Example for a proof of continuity)

We need to show that for each given $ϵ>0$ , there is a $\delta >0$ , such that for all $x\in ℝ$ with $|x-x_0|<\delta$ the inequality $|f(x)-f(x_0)|<ϵ$ holds. In our case, $x_0=1$. So by choosing $|x-1|<\delta$ for $\delta$ small enough, we may control the expression $|x-1|$ . First, let us plug $x_0=1|$ into $|f(x)-f(x_0)|$ in order to simplify the inequality to be shown $|f(x)-f(x_0)|<ϵ$ :

$\begin{array}{cc}|f(x)-f(x_0)|& =|f(x)-f(1)|\\ & =|5\cdot |x^2-2|+3-(5|1^2-2|+3)|\\ & =|5\cdot |x^2-2|-5|\\ & =5\cdot ||x^2-2|-1|\end{array}$

The objective is to "produce" as many expressions $|x-1|$ as possible, since we can control $|x-1|<\delta$. It requires some experience with epsilon-delta proofs in order to "directly see" how this is achieved. First, we need to get rid of the double absolute. This is done using the inequality $||a|-|b||\le |a-b|$ . For instance, we could use the following estimate:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots \\ & =5\cdot ||x^2-2|-1|\\ & =5\cdot ||x^2-2|-|1||\\ & \downarrow ||a|-|b||\le |a-b|\\ & \le 5\cdot |x^2-2-1|\\ & =5\cdot |x^2-3|\end{array}$

However, this is a bad estimate as the expression $|x^2-3|$ no longer tends to 0 as $x\to 1$ . To resolve this problem, we use $1=|-1|$ before applying the inequality $||a|-|b||\le |a-b|$ :

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots \\ & =5\cdot |x^2-2|-1|\\ & =5\cdot |x^2-2|-|-1||\\ & \downarrow |a-b|\ge ||a|-|b||\\ & \le 5\cdot |x^2-2-(-1)|\\ & =5\cdot |x^2-1|\end{array}$

A factor of $|x-1|$ can be directly extracted out of this with the third binomial fomula:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots \\ & \le 5\cdot |x^2-1|\\ & =5\cdot |x+1||x-1|\end{array}$

And we can control it by $|x-1|<\delta$:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots \\ & \le 5\cdot |x^2-1|\\ & =5\cdot |x+1||x-1|\\ & <5\cdot |x+1|\cdot \delta \end{array}$

Now, the required $\delta$ must only depend on $ϵ$ and $x_0$ . Therefore, we have to get rid of the $x$-dependence of $5|x+1|\cdot \delta$. This can be done by finding an upper bound for $5|x+1|$ which does not depend on $x$. As we are free to chose any $\delta$ for our proof, we may also impose any condition to it which helps us with the upper bound. In this case, $\delta \le 1$ turns out to be quite useful. In fact, $\delta \le 2$ or an even higher bound would do this job, as well. What follows from this choice?

As before, there has to be $|x-1|<\delta$. As $\delta \le 1$ , we now have $|x-1|<1$ and as $x\in [1-1;1+1]$ , we obtain $1+x\in [1;3]$ and $|x+1|\le 3$. This is the upper bound we were looking for:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots \\ & <5\cdot |x+1|\cdot \delta \\ & \downarrow |x+1|\le 3\\ & \le 15\cdot \delta \end{array}$

As we would like to show $|f(x)-f(1)|<ϵ$ , we set $\delta =\frac{ϵ}{15}$. And get that our final inequality holds:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots \\ & <15\cdot \delta \\ & \le 15\cdot \frac{ϵ}{15}\\ & \le ϵ.\end{array}$

So if the two conditions for $\delta$ are satisfied, we get the final inequality. In fact, both conditions will be satisfied if $\delta =min\{1,\frac{ϵ}{15}\}$, concluding the proof. So let's conclude our ideas and write them down in a proof:

Proof (Example for a proof of continuity)

Let $ϵ>0$ be arbitrary and let $\delta =min\{1,\frac{ϵ}{15}\}$. Further, let $x\in ℝ$ with $|x-1|<\delta$. Then:

Step 1: $|x+1|\le 3$

As $\delta =min\{1,\frac{ϵ}{15}\}$ , there is $\delta \le 1$. Hence $|x-1|\le 1$ and $x\in [1-1;1+1]$. It follows that $1+x\in [1;3]$ and therefore $|x+1|\le 3$.

Step 2: $|f(x)-f(x_0)|<ϵ$

$\begin{array}{cc}|f(x)-f(x_0)|& =|f(x)-f(1)|\\ & =|5\cdot |x^2-2|+3-(5|1^2-2|+3)|\\ & =|5\cdot |x^2-2|-5|\\ & =5\cdot ||x^2-2|-1|\\ & \downarrow 1⟺|-1|\\ & =5\cdot ||x^2-2|-|-1||\\ & \downarrow |a-b|\ge ||a|-|b||\\ & \le 5\cdot |x^2-2-(-1)|\\ & =5\cdot |x^2-1|\\ & \downarrow a^2-b^2=(a+b)(a-b)\\ & =5\cdot (|x+1|\cdot \underset{<\delta }{\underbrace{|x-1|}})\\ & <5\cdot |x+1|\cdot \delta \\ & \downarrow |x+1|\le 3\\ & \le 15\delta \\ & \downarrow \delta =\frac{ϵ}{15}\\ & =15\cdot \frac{ϵ}{15}\\ & \le ϵ\end{array}$

Hence, the function is continuous at $x_0=1$ .

Hyperbola

Exercise (Continuity of the hyperbolic function)

Prove that the function $f:{ℝ}^{+}\to ℝ$ with $f(x)=\frac{1}{x}$ is continuous.

How to get to the proof? (Continuity of the hyperbolic function)

The basic pattern for epsilon-delta proofs is applied here, as well. We would like to show the implication $|x-x_0|<\delta ⟹|f(x)-f(x_0)|<ϵ$ . Forst, let us plug in what we know and reshape our terms a bit until a $|x-x_0|$ appears:

$\begin{array}{cc}|f(x)-f(x_0)|& =|\frac{1}{x}-\frac{1}{x_0}|\\ & =\left|\frac{x-x_0}{x\cdot x_0}\right|\\ & =\frac{|x-x_0|}{|x||x_0|}\\ & =\frac{1}{|x||x_0|}\cdot |x-x_0|\end{array}$

By assumption, there will be $|x-x_0|<\delta$, of which we can make use:

$|f(x)-f(x_0)|=\dots =\frac{1}{|x||x_0|}\cdot |x-x_0|<\frac{1}{|x||x_0|}\cdot \delta$

The choice of $\delta$ may again only depend on $ϵ$ and $x_0$ so we need a smart estimate for $\frac{1}{|x|}$ in order to get rid of the $x$-dependence. To do so, we consider $\delta \le \frac{x_0}{2}$.

Why was $\delta \le \frac{x_0}{2}$ and not $\delta \le 1$ chosen? The explanation is quite simple: We need a $\delta$-neighborhood inside the domain of definition of $f$. If we had simply chosen $\delta \le 1$ , we might have get kicked out of this domain. For instance, in case$x_0=\frac{1}{2}$, the following problem appears:

The biggest $x$-value with $|x-\frac{1}{2}|<1$ is $x_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{3}{2}$ and the smallest one is $x_{\mathrm{m}\mathrm{i}\mathrm{n}}=-\frac{1}{2}$. However, $x_{\mathrm{m}\mathrm{i}\mathrm{n}}$ is not inside the domain of definition as $f:{ℝ}^{+}\to ℝ$. In particular, $x=0$ is element of that interval, where $f(x)=\frac{1}{x}$ cannot be defined at all.

A smarter choice for $\delta$, such that the $\delta$-neighborhood doesn't touch the $y$-axis is half of the distance to it, i.e. $\delta \le \frac{x_0}{2}$. A third of this distance or other fractions smaller than 1 would also be possible: $\delta \le \frac{x_0}{3}$, $\delta <\frac{x_0}{20}$ or $\delta <\frac{x_0}{5321}$.

As we chose $\delta \le \frac{x_0}{2}$ and by $|x-x_0|<\delta$ , there is $x\in [\frac{x_0}{2};\frac{3x_0}{2}]$. This allows for an upper bound: $\frac{1}{|x|}\le \frac{2}{|x_0|}$ and we may write:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots <\frac{1}{|x||x_0|}\cdot \delta \\ & \downarrow \delta =\frac{ϵ\cdot |x_0{|}^2}{2}\\ & \le \frac{2}{|x_0{|}^2}\cdot \delta \end{array}$

So we get the estimate:

$|f(x)-f(x_0)|\le \frac{2}{|x_0{|}^2}\cdot \delta$

Now, we want to prove $|f(x)-f(x_0)|<ϵ$ . Hence we choose $\delta \le \frac{ϵ\cdot |x_0{|}^2}{2}$. Plugging this in, our final inequality $|f(x)-f(x_0)|<ϵ$ will be fulfilled.

So again, we imposed two conditions for $\delta$ : $\delta \le \frac{x_0}{2}$ and $\delta \le \frac{ϵ\cdot |x_0{|}^2}{2}$. Both are fulfilled by $\delta =min\{\frac{x_0}{2},\frac{ϵ\cdot |x_0{|}^2}{2}\}$, which we will use in the proof:

Proof (Continuity of the hyperbolic function)

Let $f:{ℝ}^{+}\to ℝ$ with $f(x)=\frac{1}{x}$ and let $x_0\in {ℝ}^{+}$ be arbitrary. Further, let $ϵ>0$ be arbitrary. We choose $\delta :=min\{\frac{x_0}{2},\frac{ϵ\cdot |x_0{|}^2}{2}\}$. For all $x\in ℝ$ with $|x-x_0|<\delta$ there is:

$\begin{array}{cc}|f(x)-f(x_0)|& =|\frac{1}{x}-\frac{1}{x_0}|\\ & =\left|\frac{x-x_0}{x\cdot x_0}\right|\\ & =\frac{|x-x_0|}{|x||x_0|}\\ & =\frac{1}{|x||x_0|}|x-x_0|\\ & \downarrow \frac{1}{|x|}\le \frac{2}{|x_0|}\\ & \le \frac{2}{|x_0{|}^2}|x-x_0|\\ & \downarrow |x-x_0|<\delta \\ & <\frac{2\delta }{|x_0{|}^2}\\ & \downarrow \delta =\frac{ϵ\cdot |x_0{|}^2}{2}\\ & =\frac{2}{|x_0{|}^2}\cdot \frac{ϵ\cdot |x_0{|}^2}{2}\\ & =ϵ\end{array}$

Hence, the function $f$ is continuous at $x_0$ . And as $x_0$ was chosen to be arbitrary, the whole function $f$ is continuous.

Concatenated square root function

Exercise (Epsilon-Delta proof for the continuity of the Square Root Function)

Show, using the epsilon-delta criterion, that the following function is continuous:

$f:ℝ\to ℝ,x\mapsto \sqrt{5+x^2}$

How to get to the proof? (Epsilon-Delta proof for the continuity of the Square Root Function)

We need to show, that for any given $ϵ>0$ , there is a $\delta >0$ , such that all $x\in ℝ$ with $|x-a|<\delta$ satisfy the inequality $|f(x)-f(a)|<ϵ$ . So let us take a look at the target inequality $|f(x)-f(a)|<ϵ$ and estimate the absolute $|f(x)-f(a)|$ from above. We are able to control the term $|x-a|$ . Therefor, would like to get an upper bound for $|f(x)-f(a)|$ including the expression $|x-a|$ . So we are looking for an inequality of the form

$|f(x)-f(a)|\le K(x,a)\cdot |x-a|$

Here, $K(x,a)$ is some expression depending on $x$ and $a$ . The second factor is smaller than $\delta$ and can be made arbitrarily small by a suitable choice of $\delta$ . Such a bound is constructed as follows:

$\begin{array}{cc}|f(x)-f(a)|& =|\sqrt{5+x^2}-\sqrt{5+a^2}|\\ & \downarrow \text{expand with}|\sqrt{5+x^2}+\sqrt{5+a^2}|\\ & =\frac{|\sqrt{5+x^2}-\sqrt{5+a^2}|\cdot |\sqrt{5+x^2}+\sqrt{5+a^2}|}{|\sqrt{5+x^2}+\sqrt{5+a^2}|}\\ & \downarrow \sqrt{5+x^2}+\sqrt{5+a^2}\ge 0\\ & =\frac{|x^2-a^2|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & =\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\cdot |x-a|\\ & \downarrow K(x,a):=\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & =K(x,a)\cdot |x-a|\end{array}$

Since $|x-a|<\delta$ , there is:

$|f(x)-f(a)|=K(x,a)\cdot |x-a|<K(x,a)\cdot \delta$

If we now choose $\delta$ small enough, such that $K(x,a)\cdot \delta \le ϵ$ , then we obtain our target inequality $|f(x)-f(a)|\le ϵ$. But $K(x,a)$ still depends on $x$ , so $\delta$ would have to depend on , too - and we required one choice of $\delta$ which is suitable for all $x$ . Therefore, we need to get rid of the $x$-dependence. This is done by an estimate of the first factor, such that our inequality takes the form $K(x,a)\le \tilde{K}(a)$ :

$\begin{array}{cc}K(x,a)& =\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & \downarrow \text{triangle inequality}\\ & \le \frac{\left|x\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}}+\frac{\left|a\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & \downarrow \frac{a}{b+c}\le \frac{a}{b}\text{ for }a,c\ge 0,b>0\\ & \le \frac{\left|x\right|}{\sqrt{5+x^2}}+\frac{\left|a\right|}{\sqrt{5+a^2}}\\ & \downarrow \frac{|a|}{\sqrt{5+a^2}}\le 1\text{, since }\sqrt{5+a^2}\ge \sqrt{a^2}=|a|\\ & \le 2=:\tilde{K}(a)\end{array}$

We even made $\tilde{K}(a)$ independent of $a$ , which would in fact not have been necessary. So we obtain the following inequality

$|f(x)-f(a)|\le 2\cdot |x-a|<2\cdot \delta$

We need the estimate $2\cdot \delta \le ϵ$, in order to fulfill the target inequality $|f(x)-f(a)|<ϵ$ . The choice of $\delta =\frac{ϵ}{2}$ is sufficient for that. So let us write down the proof:

Proof (Epsilon-Delta proof for the continuity of the Square Root Function)

Let $f:ℝ\to ℝ$ with $f(x)=\sqrt{5+x^2}$. Let $a\in ℝ$ and an arbitrary $ϵ>0$ be given. We choose $\delta =\frac{ϵ}{2}$. For all $x\in ℝ$ with $|x-a|<\delta$ there is:

$\begin{array}{cc}|f(x)-f(a)|& =|\sqrt{5+x^2}-\sqrt{5+a^2}|\\ & =\frac{|\sqrt{5+x^2}-\sqrt{5+a^2}|\cdot |\sqrt{5+x^2}+\sqrt{5+a^2}|}{|\sqrt{5+x^2}+\sqrt{5+a^2}|}\\ & =\frac{|x^2-a^2|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & =\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\cdot |x-a|\\ & \le (\frac{\left|x\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}}+\frac{\left|a\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}})\cdot |x-a|\\ & \le (\frac{\left|x\right|}{\sqrt{5+x^2}}+\frac{\left|a\right|}{\sqrt{5+a^2}})\cdot |x-a|\\ & \le (1+1)\cdot |x-a|\\ & \le 2\cdot |x-a|\\ & \downarrow |x-a|<\delta =\frac{ϵ}{2}\\ & <2\cdot \frac{ϵ}{2}=ϵ\end{array}$

Hence, $f$ is a continuous function.

Discontinuity of the topological sine function

Exercise (Discontinuity of the topological sine fucntion)

Prove the discontinuity at $x_0=0$ for the topological sine function:

$f:ℝ\to ℝ:x\mapsto \{\begin{array}{cc}\sin \left(\frac{1}{x}\right)& x\ne 0\\ 0& x=0\end{array}$

How to get to the proof? (Discontinuity of the topological sine fucntion)

In this exercise, discontinuity has to be shown for a given function. This is done by the negation of the epsilon-delta criterion. Our objective is to find both an $ϵ>0$ and an $x\in ℝ$, such that $|x-x_0|<\delta$ and $|f(x)-f(x_0)|\ge ϵ$ . Here, $x$ may be chosen depending on $\delta$ , while $ϵ$ has to be the same for all $\delta >0$ . For a solution, we may proceed as follows

Step 1: Simplify the target inequality

First, we may simplify the two inequalities which have to be fulfilled by plugging in $x_0=0$ and $f(x_0)=0$. We therefore get: $|x|<\delta$ and $|f(x)|\ge ϵ$.

Step 2: Choose a suitable $ϵ>0$

We consider the graph of the function $f$ . It will help us finding the building bricks for our proof:

Graph of the topological sine function

We need to find an $ϵ>0$ , such that there are arguments in each arbitrarily narrow interval $(x_0-\delta ,x_0+\delta )=(-\delta ,\delta )$ whose function values have a distance larger than $ϵ$ from $f(x_0)=f(0)=0$ . Visually, no matter how small the width $\delta$ of the $2ϵ$-$2\delta$-rectangle is chosen, there will always be some points below or above it.

Taking a look at the graph, we see that our function oscillates between $-1$ and $1$ . Hence, $ϵ\le 1$ may be useful. In that case, there will be function values with $|f(x)|\ge 1$ in every arbitrarily small neighborhood around $x_0=0$. We choose $ϵ=\frac{1}{2}$. This is visualized in the following figure:

The topological sine function oscillates between -1 and 1 nea the origin. Hence, in each neighborhood around 0, there will be arguments x , where f(x) has a distance larger than 1/2 to 0.

After $ϵ$ has been chosen, an arbitrary $\delta >0$ will be assumed, for which we need to find a suitable $x$. This is what we will do next.

Step 3: Choice of a suitable $x\in ℝ$

We just set $ϵ=\frac{1}{2}$ . Therefore, $|\sin \left(\frac{1}{x}\right)|\ge \frac{1}{2}$ has to hold. So it would be nice to choose an $x$ with $\sin \left(\frac{1}{x}\right)=1$ . Now, $\sin (a)=1$ is obtained for any $a=\frac{\pi }{2}+2k\pi$ with $k\in \mathbb{Z}$ . The condition for $x$ such that the function gets 1 is therefore:

$\frac{1}{x}=\frac{\pi }{2}+2k\pi ⟺x=\frac{1}{\frac{\pi }{2}+2k\pi }$

So we found several $x$ , where $|f(x)|\ge ϵ$ . Now we only need to select one among them, which satisfies $|x|<\delta$ for the given $\delta$ . Our $x$ depend on $k$ . So we have to select a suitable $k\in \mathbb{Z}$ , where $|x|<\delta$ . To do so, let us plug $x=\frac{1}{\frac{\pi }{2}+2k\pi }$ into this inequality and solve it for $k$ :

$\begin{array}{cc}\left|x\right|<\delta & ⟺\left|\frac{1}{\frac{\pi }{2}+2k\pi }\right|<\delta \\ & ⟺\frac{1}{\frac{\pi }{2}+2k\pi }<\delta \\ & ⟺2k\pi +\frac{\pi }{2}>\frac{1}{\delta }\\ & ⟺2k\pi >\frac{1}{\delta }-\frac{\pi }{2}\\ & ⟺k>\frac{1}{2\pi }(\frac{1}{\delta }-\frac{\pi }{2})\end{array}$

So the condition on $k$ is $k>\frac{1}{2\pi }(\frac{1}{\delta }-\frac{\pi }{2})$. If we choose just any natural number $k$ above this threshold $\frac{1}{2\pi }(\frac{1}{\delta }-\frac{\pi }{2})$ , then $|x|<\delta$ will be fulfilled. Such a $k$ has to exist by Archimedes' axiom (for instance by flooring up the right-hand expression). So let us choose such a $k$ and define $x$ via $x=\frac{1}{\frac{\pi }{2}+2k\pi }$ . This gives us both $|x|<\delta$ and $|f(x)|\ge ϵ$ . So we got all building bricks together, which we will now assemble to a final proof:

Proof (Discontinuity of the topological sine fucntion)

Choose $ϵ=\frac{1}{2}$ and let $\delta >0$ be arbitrary. Choose a natural number $k$ with $k>\frac{1}{2\pi }(\frac{1}{\delta }-\frac{\pi }{2})$. Such a natural number $k$ has to exist by Archimedes' axiom. Further, let $x=\frac{1}{\frac{\pi }{2}+2k\pi }$. Then:

$\begin{array}{cc}k>\frac{1}{2\pi }(\frac{1}{\delta }-\frac{\pi }{2})& ⟹2k\pi >\frac{1}{\delta }-\frac{\pi }{2}\\ & ⟹2k\pi +\frac{\pi }{2}>\frac{1}{\delta }\\ & \downarrow \frac{1}{a}>\frac{1}{b}>0⟺0<a<b\\ & ⟹\frac{1}{\frac{\pi }{2}+2k\pi }<\delta \\ & \downarrow \frac{1}{\frac{\pi }{2}+2k\pi }>0\\ & ⟹\left|\frac{1}{\frac{\pi }{2}+2k\pi }\right|<\delta \\ & \downarrow x=\frac{1}{\frac{\pi }{2}+2k\pi }\\ & ⟹\left|x\right|<\delta \end{array}$

In addition:

$\begin{array}{cc}|f(x)-f(0)|& =|\sin \left(\frac{1}{x}\right)-0|\\ & \downarrow x=\frac{1}{\frac{\pi }{2}+2k\pi }\\ & =|\sin (\frac{\pi }{2}+2k\pi )|\\ & \downarrow \sin (\frac{\pi }{2}+2k\pi )=1\\ & =\left|1\right|\ge \frac{1}{2}=ϵ\end{array}$

Hence, the function is discontinuous at $x_0=0$.

Example Problems

Sequence Criterion: Absolute Value Function

Exercise (Continuity of the absolute function)

Prove continuity for the absolute function.

Proof (Continuity of the absolute function)

Let $f:ℝ\to ℝ$ with $f(x)=|x|$ be the absolute function. Let $x_0\in ℝ$ be a real number and $(x_n{)}_{n\in \mathbb{N}}$ a sequence converging to it $\underset{n\to \mathrm{\infty }}{lim}{x}_n=x_0$. In the chapter on limit theorems we have proven the absolute rule, stating that $\underset{n\to \mathrm{\infty }}{lim}|x_n|=|x_0|$ , whenever there is $\underset{n\to \mathrm{\infty }}{lim}{x}_n=x_0$ . Hence:

$\underset{n\to \mathrm{\infty }}{lim}f(x_n)=\underset{n\to \mathrm{\infty }}{lim}|x_n|=|x_0|=f(x_0)$

This proves continuity of the absolute function $f$ by the sequence criterion.

Discontinuity of the topological sine function

Exercise (Discontinuity of the topological sine function)

Prove discontinuity of the following function:

$f:ℝ\to ℝ:x\mapsto \{\begin{array}{cc}\sin \left(\frac{1}{x}\right)& x\ne 0\\ 0& x=0\end{array}$

How to get to the proof? (Discontinuity of the topological sine function)

Discontinuity of $f$ means that this function has at least one argument where $f$ is discontinuous. For each $x\ne 0$ , $f$ is equal to the function $\sin \left(\frac{1}{x}\right)$ in a sufficiently small neighbourhood of $x$. Since $\sin \left(\frac{1}{x}\right)$ is just a composition of continuous functions, it is continuous itself and therefore, $f$ must be continuous for all $x\ne 0$ , as well. So we know that the discontinuity may only be situated at $x=0$ .

In order to prove that $f$ is discontinuous at $x=0$ , we need to find a sequence of arguments $(x_n{)}_{n\in \mathbb{N}}$ converging to $\underset{n\to \mathrm{\infty }}{lim}{x}_n=0$ but with $\underset{n\to \mathrm{\infty }}{lim}f(x_n)\ne f(0)$ . To find such a function, let us take a look at the graph of the function $f$ :

Graph of the topological sine function f

In this figure, we recognize that $f$ takes any value between $-1$ and $1$ infinitely often in the vicinity of $x=0$. So, for instance, we may just choose $(x_n{)}_{n\in \mathbb{N}}$ such that $f(x_n)$ is always $1$ . This guarantees that $\underset{n\to \mathrm{\infty }}{lim}f(x_n)=1\ne 0=f(0)$ - and actually any other real number $f(x_n)\ne 0$ between $-1$ and $1$ in place of $f(x_n)=1$ would do the job. But we need to make a specific choice for $x_n$ , and $f(x_n)=1$ is a very simple one. In addition, we will choose $x_n$ to converges to zero from above.

The following figure also contains the sequence elements $x_n$ beside our function $f$. We may clearly see that for $x_n\to 0$ the sequence of function values converges to $1$ , which is different from the function value $f(0)=0$ :

Graph of the topological sine function f together with the function values for our sequence of arguments, which converges to 0 from the right and always takes function value 1.

But what are the exact values of these $x_n$ for which we would like to have $f(x_n)=1$ To answer this question, let us resolve the equation $f(x)=1$ for $x$ :

$\begin{array}{c}\begin{array}{cccc}& & f(x)& =1\\ \stackrel{f(0)\ne 0}{⟺}& & \sin \left(\frac{1}{x}\right)& =1\\ ⟺& \mathrm{\exists }k\in \mathbb{Z}:& \frac{1}{x}& =\frac{\pi }{2}+2k\pi \\ ⟺& \mathrm{\exists }k\in \mathbb{Z}:& x& =\frac{1}{\frac{\pi }{2}+2k\pi }\end{array}\end{array}$

So for each $x=\frac{1}{\frac{\pi }{2}+2k\pi }$ with $k\in \mathbb{Z}$ , we have $f(x)=1$. In order to get positive $x_n$ converging to zero from above, we may for instance choose $x_n=\frac{1}{\frac{\pi }{2}+2n\pi }$. In that case:

$\underset{n\to \mathrm{\infty }}{lim}{x}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\frac{\pi }{2}+2n\pi }=0$

And we have seen that $\underset{n\to \mathrm{\infty }}{lim}f(x_n)=1\ne 0=f(0)$ . So we found just a sequence of arguments $(x_n{)}_{n\in \mathbb{N}}$ , which proves discontinuity of $f$ at $x=0$ .

Proof (Discontinuity of the topological sine function)

Let $f:ℝ\to ℝ$ with $f(x)=\sin \left(\frac{1}{x}\right)$ for $x\ne 0$ and $f(0)=0$. We consider the sequence $(x_n{)}_{n\in \mathbb{N}}$ defined by $x_n=\frac{1}{\frac{\pi }{2}+2n\pi }$. For this sequence:

$\underset{n\to \mathrm{\infty }}{lim}{x}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\frac{\pi }{2}+2n\pi }=0$

And there is:

$\begin{array}{cc}\underset{n\to \mathrm{\infty }}{lim}f(x_n)& =\underset{n\to \mathrm{\infty }}{lim}f\left(\frac{1}{\frac{\pi }{2}+2n\pi }\right)\\ & =\underset{n\to \mathrm{\infty }}{lim}\sin \left(\frac{1}{\frac{1}{\frac{\pi }{2}+2n\pi }}\right)\\ & =\underset{n\to \mathrm{\infty }}{lim}\sin (\frac{\pi }{2}+2n\pi )\\ & =\underset{n\to \mathrm{\infty }}{lim}1=1\end{array}$

Hence, $\underset{n\to \mathrm{\infty }}{lim}f(x_n)=1\ne 0=f(0)$ although $\underset{n\to \mathrm{\infty }}{lim}{x}_n=0$ . This proves that $f$ is discontinuous at $x=0$ and therefore it is a discontinuous function.

Epsilon-Delta Criterion: Linear Function

Exercise (Continuity of a linear function)

Prove that a linear function $f:ℝ\to ℝ$ with $f(x)=\frac{1}{3}x$ is continuous.

How to get to the proof? (Continuity of a linear function)

Graph of a function $f:ℝ\to ℝ$ with $f(x)=\frac{1}{3}x$. Considering the graph , we see that this function is continuous everywhere.

To actually prove continuity of $f$ , we need to check continuity at any argument $x_0\in ℝ$ . So let $x_0$ be an arbitrary real number. Now, choose any arbitrary maximal error $ϵ>0$ . Our task is now to find a sufficiently small $\delta >0$ , such that $|f(x)-f(x_0)|<ϵ$ for all arguments $x$ with $|x-x_0|<\delta$ . Let us take a closer look at the inequality $|f(x)-f(x_0)|<ϵ$ :

$\begin{array}{c}\begin{array}{ccc}& |f(x)-f(x_0)|& <ϵ\\ ⟺& |\frac{1}{3}x-\frac{1}{3}{x}_0|& <ϵ\\ ⟺& |\frac{1}{3}(x-x_0)|& <ϵ\\ ⟺& \frac{1}{3}|x-x_0|& <ϵ\end{array}\end{array}$

That means, $\frac{1}{3}|x-x_0|<ϵ$ has to be fulfilled for all $x$ with $|x-x_0|<\delta$ . How to choose $\delta$ , such that $|x-x_0|<\delta$ implies $\frac{1}{3}|x-x_0|<ϵ$ ?

We use that the inequality $\frac{1}{3}|x-x_0|<ϵ$ contains the distance $|x-x_0|$ . As $|x-x_0|<\delta$ we know that this distance is smaller than $\delta$ . This can be plugged into the inequality $\frac{1}{3}|x-x_0|$ :

$\frac{1}{3}|x-x_0|\stackrel{|x-x_0|<\delta }{<}\frac{1}{3}\delta$

If $\delta$ is now chosen such that $\frac{1}{3}\delta \le ϵ$ , then $\frac{1}{3}|x-x_0|<\frac{1}{3}\delta$ will yield the inequality $\frac{1}{3}|x-x_0|<ϵ$ which we wanted to show. The smallness condition for $\delta$ can now simply be found by resolving $\frac{1}{3}\delta \le ϵ$ for $\delta$:

$\frac{1}{3}\delta \le ϵ⟺\delta \le 3ϵ$

Any $\delta$ satisfying $0<\delta \le 3ϵ$ could be used for the proof. For instance, we may use $\delta =3ϵ$. As we now found a suitable $\delta$, we can finally conduct the proof:

Proof (Continuity of a linear function)

Let $f:ℝ\to ℝ$ with $f(x)=\frac{1}{3}x$ and let $x_0\in ℝ$ be arbitrary. In addition, consider any $ϵ>0$ to be given. We choose $\delta =3ϵ$. Let $x\in ℝ$ with $|x-x_0|<\delta$. There is:

$\begin{array}{cc}|f(x)-f(x_0)|& =|\frac{1}{3}x-\frac{1}{3}{x}_0|\\ & =|\frac{1}{3}(x-x_0)|\\ & =\frac{1}{3}|x-x_0|\\ & \downarrow |x-x_0|<\delta \\ & <\frac{1}{3}\delta \\ & \downarrow \delta =3ϵ\\ & \le \frac{1}{3}(3ϵ)=ϵ\end{array}$

This shows $|f(x)-f(x_0)|<ϵ$, and establishes continuity of $f$ at $x_0$ by means of the epsilon-delta criterion. Since $x_0\in ℝ$ was chosen to be arbitrary, we also know that the entire function $f$ is continuous.

Epsilon-Delta Criterion: Concatenated Absolute Value Function

Exercise (Example for a proof of continuity)

Prove that the following function is continuous at $x_0=1$:

$f:ℝ\to ℝ:x\mapsto f(x)=5|x^2-2|+3$

How to get to the proof? (Example for a proof of continuity)

We need to show that for each given $ϵ>0$ , there is a $\delta >0$ , such that for all $x\in ℝ$ with $|x-x_0|<\delta$ the inequality $|f(x)-f(x_0)|<ϵ$ holds. In our case, $x_0=1$. So by choosing $|x-1|<\delta$ for $\delta$ small enough, we may control the expression $|x-1|$ . First, let us plug $x_0=1|$ into $|f(x)-f(x_0)|$ in order to simplify the inequality to be shown $|f(x)-f(x_0)|<ϵ$ :

$\begin{array}{cc}|f(x)-f(x_0)|& =|f(x)-f(1)|\\ & =|5\cdot |x^2-2|+3-(5|1^2-2|+3)|\\ & =|5\cdot |x^2-2|-5|\\ & =5\cdot ||x^2-2|-1|\end{array}$

The objective is to "produce" as many expressions $|x-1|$ as possible, since we can control $|x-1|<\delta$. It requires some experience with epsilon-delta proofs in order to "directly see" how this is achieved. First, we need to get rid of the double absolute. This is done using the inequality $||a|-|b||\le |a-b|$ . For instance, we could use the following estimate:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots \\ & =5\cdot ||x^2-2|-1|\\ & =5\cdot ||x^2-2|-|1||\\ & \downarrow ||a|-|b||\le |a-b|\\ & \le 5\cdot |x^2-2-1|\\ & =5\cdot |x^2-3|\end{array}$

However, this is a bad estimate as the expression $|x^2-3|$ no longer tends to 0 as $x\to 1$ . To resolve this problem, we use $1=|-1|$ before applying the inequality $||a|-|b||\le |a-b|$ :

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots \\ & =5\cdot |x^2-2|-1|\\ & =5\cdot |x^2-2|-|-1||\\ & \downarrow |a-b|\ge ||a|-|b||\\ & \le 5\cdot |x^2-2-(-1)|\\ & =5\cdot |x^2-1|\end{array}$

A factor of $|x-1|$ can be directly extracted out of this with the third binomial fomula:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots \\ & \le 5\cdot |x^2-1|\\ & =5\cdot |x+1||x-1|\end{array}$

And we can control it by $|x-1|<\delta$:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots \\ & \le 5\cdot |x^2-1|\\ & =5\cdot |x+1||x-1|\\ & <5\cdot |x+1|\cdot \delta \end{array}$

Now, the required $\delta$ must only depend on $ϵ$ and $x_0$ . Therefore, we have to get rid of the $x$-dependence of $5|x+1|\cdot \delta$. This can be done by finding an upper bound for $5|x+1|$ which does not depend on $x$. As we are free to chose any $\delta$ for our proof, we may also impose any condition to it which helps us with the upper bound. In this case, $\delta \le 1$ turns out to be quite useful. In fact, $\delta \le 2$ or an even higher bound would do this job, as well. What follows from this choice?

As before, there has to be $|x-1|<\delta$. As $\delta \le 1$ , we now have $|x-1|<1$ and as $x\in [1-1;1+1]$ , we obtain $1+x\in [1;3]$ and $|x+1|\le 3$. This is the upper bound we were looking for:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots \\ & <5\cdot |x+1|\cdot \delta \\ & \downarrow |x+1|\le 3\\ & \le 15\cdot \delta \end{array}$

As we would like to show $|f(x)-f(1)|<ϵ$ , we set $\delta =\frac{ϵ}{15}$. And get that our final inequality holds:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots \\ & <15\cdot \delta \\ & \le 15\cdot \frac{ϵ}{15}\\ & \le ϵ.\end{array}$

So if the two conditions for $\delta$ are satisfied, we get the final inequality. In fact, both conditions will be satisfied if $\delta =min\{1,\frac{ϵ}{15}\}$, concluding the proof. So let's conclude our ideas and write them down in a proof:

Proof (Example for a proof of continuity)

Let $ϵ>0$ be arbitrary and let $\delta =min\{1,\frac{ϵ}{15}\}$. Further, let $x\in ℝ$ with $|x-1|<\delta$. Then:

Step 1: $|x+1|\le 3$

As $\delta =min\{1,\frac{ϵ}{15}\}$ , there is $\delta \le 1$. Hence $|x-1|\le 1$ and $x\in [1-1;1+1]$. It follows that $1+x\in [1;3]$ and therefore $|x+1|\le 3$.

Step 2: $|f(x)-f(x_0)|<ϵ$

$\begin{array}{cc}|f(x)-f(x_0)|& =|f(x)-f(1)|\\ & =|5\cdot |x^2-2|+3-(5|1^2-2|+3)|\\ & =|5\cdot |x^2-2|-5|\\ & =5\cdot ||x^2-2|-1|\\ & \downarrow 1⟺|-1|\\ & =5\cdot ||x^2-2|-|-1||\\ & \downarrow |a-b|\ge ||a|-|b||\\ & \le 5\cdot |x^2-2-(-1)|\\ & =5\cdot |x^2-1|\\ & \downarrow a^2-b^2=(a+b)(a-b)\\ & =5\cdot (|x+1|\cdot \underset{<\delta }{\underbrace{|x-1|}})\\ & <5\cdot |x+1|\cdot \delta \\ & \downarrow |x+1|\le 3\\ & \le 15\delta \\ & \downarrow \delta =\frac{ϵ}{15}\\ & =15\cdot \frac{ϵ}{15}\\ & \le ϵ\end{array}$

Hence, the function is continuous at $x_0=1$ .

Hyperbola

Exercise (Continuity of the hyperbolic function)

Prove that the function $f:{ℝ}^{+}\to ℝ$ with $f(x)=\frac{1}{x}$ is continuous.

How to get to the proof? (Continuity of the hyperbolic function)

The basic pattern for epsilon-delta proofs is applied here, as well. We would like to show the implication $|x-x_0|<\delta ⟹|f(x)-f(x_0)|<ϵ$ . Forst, let us plug in what we know and reshape our terms a bit until a $|x-x_0|$ appears:

$\begin{array}{cc}|f(x)-f(x_0)|& =|\frac{1}{x}-\frac{1}{x_0}|\\ & =\left|\frac{x-x_0}{x\cdot x_0}\right|\\ & =\frac{|x-x_0|}{|x||x_0|}\\ & =\frac{1}{|x||x_0|}\cdot |x-x_0|\end{array}$

By assumption, there will be $|x-x_0|<\delta$, of which we can make use:

$|f(x)-f(x_0)|=\dots =\frac{1}{|x||x_0|}\cdot |x-x_0|<\frac{1}{|x||x_0|}\cdot \delta$

The choice of $\delta$ may again only depend on $ϵ$ and $x_0$ so we need a smart estimate for $\frac{1}{|x|}$ in order to get rid of the $x$-dependence. To do so, we consider $\delta \le \frac{x_0}{2}$.

Why was $\delta \le \frac{x_0}{2}$ and not $\delta \le 1$ chosen? The explanation is quite simple: We need a $\delta$-neighborhood inside the domain of definition of $f$. If we had simply chosen $\delta \le 1$ , we might have get kicked out of this domain. For instance, in case$x_0=\frac{1}{2}$, the following problem appears:

The biggest $x$-value with $|x-\frac{1}{2}|<1$ is $x_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{3}{2}$ and the smallest one is $x_{\mathrm{m}\mathrm{i}\mathrm{n}}=-\frac{1}{2}$. However, $x_{\mathrm{m}\mathrm{i}\mathrm{n}}$ is not inside the domain of definition as $f:{ℝ}^{+}\to ℝ$. In particular, $x=0$ is element of that interval, where $f(x)=\frac{1}{x}$ cannot be defined at all.

A smarter choice for $\delta$, such that the $\delta$-neighborhood doesn't touch the $y$-axis is half of the distance to it, i.e. $\delta \le \frac{x_0}{2}$. A third of this distance or other fractions smaller than 1 would also be possible: $\delta \le \frac{x_0}{3}$, $\delta <\frac{x_0}{20}$ or $\delta <\frac{x_0}{5321}$.

As we chose $\delta \le \frac{x_0}{2}$ and by $|x-x_0|<\delta$ , there is $x\in [\frac{x_0}{2};\frac{3x_0}{2}]$. This allows for an upper bound: $\frac{1}{|x|}\le \frac{2}{|x_0|}$ and we may write:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots <\frac{1}{|x||x_0|}\cdot \delta \\ & \downarrow \delta =\frac{ϵ\cdot |x_0{|}^2}{2}\\ & \le \frac{2}{|x_0{|}^2}\cdot \delta \end{array}$

So we get the estimate:

$|f(x)-f(x_0)|\le \frac{2}{|x_0{|}^2}\cdot \delta$

Now, we want to prove $|f(x)-f(x_0)|<ϵ$ . Hence we choose $\delta \le \frac{ϵ\cdot |x_0{|}^2}{2}$. Plugging this in, our final inequality $|f(x)-f(x_0)|<ϵ$ will be fulfilled.

So again, we imposed two conditions for $\delta$ : $\delta \le \frac{x_0}{2}$ and $\delta \le \frac{ϵ\cdot |x_0{|}^2}{2}$. Both are fulfilled by $\delta =min\{\frac{x_0}{2},\frac{ϵ\cdot |x_0{|}^2}{2}\}$, which we will use in the proof:

Proof (Continuity of the hyperbolic function)

Let $f:{ℝ}^{+}\to ℝ$ with $f(x)=\frac{1}{x}$ and let $x_0\in {ℝ}^{+}$ be arbitrary. Further, let $ϵ>0$ be arbitrary. We choose $\delta :=min\{\frac{x_0}{2},\frac{ϵ\cdot |x_0{|}^2}{2}\}$. For all $x\in ℝ$ with $|x-x_0|<\delta$ there is:

$\begin{array}{cc}|f(x)-f(x_0)|& =|\frac{1}{x}-\frac{1}{x_0}|\\ & =\left|\frac{x-x_0}{x\cdot x_0}\right|\\ & =\frac{|x-x_0|}{|x||x_0|}\\ & =\frac{1}{|x||x_0|}|x-x_0|\\ & \downarrow \frac{1}{|x|}\le \frac{2}{|x_0|}\\ & \le \frac{2}{|x_0{|}^2}|x-x_0|\\ & \downarrow |x-x_0|<\delta \\ & <\frac{2\delta }{|x_0{|}^2}\\ & \downarrow \delta =\frac{ϵ\cdot |x_0{|}^2}{2}\\ & =\frac{2}{|x_0{|}^2}\cdot \frac{ϵ\cdot |x_0{|}^2}{2}\\ & =ϵ\end{array}$

Hence, the function $f$ is continuous at $x_0$ . And as $x_0$ was chosen to be arbitrary, the whole function $f$ is continuous.

Concatenated square root function

Exercise (Epsilon-Delta proof for the continuity of the Square Root Function)

Show, using the epsilon-delta criterion, that the following function is continuous:

$f:ℝ\to ℝ,x\mapsto \sqrt{5+x^2}$

How to get to the proof? (Epsilon-Delta proof for the continuity of the Square Root Function)

We need to show, that for any given $ϵ>0$ , there is a $\delta >0$ , such that all $x\in ℝ$ with $|x-a|<\delta$ satisfy the inequality $|f(x)-f(a)|<ϵ$ . So let us take a look at the target inequality $|f(x)-f(a)|<ϵ$ and estimate the absolute $|f(x)-f(a)|$ from above. We are able to control the term $|x-a|$ . Therefor, would like to get an upper bound for $|f(x)-f(a)|$ including the expression $|x-a|$ . So we are looking for an inequality of the form

$|f(x)-f(a)|\le K(x,a)\cdot |x-a|$

Here, $K(x,a)$ is some expression depending on $x$ and $a$ . The second factor is smaller than $\delta$ and can be made arbitrarily small by a suitable choice of $\delta$ . Such a bound is constructed as follows:

$\begin{array}{cc}|f(x)-f(a)|& =|\sqrt{5+x^2}-\sqrt{5+a^2}|\\ & \downarrow \text{expand with}|\sqrt{5+x^2}+\sqrt{5+a^2}|\\ & =\frac{|\sqrt{5+x^2}-\sqrt{5+a^2}|\cdot |\sqrt{5+x^2}+\sqrt{5+a^2}|}{|\sqrt{5+x^2}+\sqrt{5+a^2}|}\\ & \downarrow \sqrt{5+x^2}+\sqrt{5+a^2}\ge 0\\ & =\frac{|x^2-a^2|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & =\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\cdot |x-a|\\ & \downarrow K(x,a):=\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & =K(x,a)\cdot |x-a|\end{array}$

Since $|x-a|<\delta$ , there is:

$|f(x)-f(a)|=K(x,a)\cdot |x-a|<K(x,a)\cdot \delta$

If we now choose $\delta$ small enough, such that $K(x,a)\cdot \delta \le ϵ$ , then we obtain our target inequality $|f(x)-f(a)|\le ϵ$. But $K(x,a)$ still depends on $x$ , so $\delta$ would have to depend on , too - and we required one choice of $\delta$ which is suitable for all $x$ . Therefore, we need to get rid of the $x$-dependence. This is done by an estimate of the first factor, such that our inequality takes the form $K(x,a)\le \tilde{K}(a)$ :

$\begin{array}{cc}K(x,a)& =\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & \downarrow \text{triangle inequality}\\ & \le \frac{\left|x\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}}+\frac{\left|a\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & \downarrow \frac{a}{b+c}\le \frac{a}{b}\text{ for }a,c\ge 0,b>0\\ & \le \frac{\left|x\right|}{\sqrt{5+x^2}}+\frac{\left|a\right|}{\sqrt{5+a^2}}\\ & \downarrow \frac{|a|}{\sqrt{5+a^2}}\le 1\text{, since }\sqrt{5+a^2}\ge \sqrt{a^2}=|a|\\ & \le 2=:\tilde{K}(a)\end{array}$

We even made $\tilde{K}(a)$ independent of $a$ , which would in fact not have been necessary. So we obtain the following inequality

$|f(x)-f(a)|\le 2\cdot |x-a|<2\cdot \delta$

We need the estimate $2\cdot \delta \le ϵ$, in order to fulfill the target inequality $|f(x)-f(a)|<ϵ$ . The choice of $\delta =\frac{ϵ}{2}$ is sufficient for that. So let us write down the proof:

Proof (Epsilon-Delta proof for the continuity of the Square Root Function)

Let $f:ℝ\to ℝ$ with $f(x)=\sqrt{5+x^2}$. Let $a\in ℝ$ and an arbitrary $ϵ>0$ be given. We choose $\delta =\frac{ϵ}{2}$. For all $x\in ℝ$ with $|x-a|<\delta$ there is:

$\begin{array}{cc}|f(x)-f(a)|& =|\sqrt{5+x^2}-\sqrt{5+a^2}|\\ & =\frac{|\sqrt{5+x^2}-\sqrt{5+a^2}|\cdot |\sqrt{5+x^2}+\sqrt{5+a^2}|}{|\sqrt{5+x^2}+\sqrt{5+a^2}|}\\ & =\frac{|x^2-a^2|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & =\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\cdot |x-a|\\ & \le (\frac{\left|x\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}}+\frac{\left|a\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}})\cdot |x-a|\\ & \le (\frac{\left|x\right|}{\sqrt{5+x^2}}+\frac{\left|a\right|}{\sqrt{5+a^2}})\cdot |x-a|\\ & \le (1+1)\cdot |x-a|\\ & \le 2\cdot |x-a|\\ & \downarrow |x-a|<\delta =\frac{ϵ}{2}\\ & <2\cdot \frac{ϵ}{2}=ϵ\end{array}$

Hence, $f$ is a continuous function.

Discontinuity of the topological sine function

Exercise (Discontinuity of the topological sine fucntion)

Prove the discontinuity at $x_0=0$ for the topological sine function:

$f:ℝ\to ℝ:x\mapsto \{\begin{array}{cc}\sin \left(\frac{1}{x}\right)& x\ne 0\\ 0& x=0\end{array}$

How to get to the proof? (Discontinuity of the topological sine fucntion)

In this exercise, discontinuity has to be shown for a given function. This is done by the negation of the epsilon-delta criterion. Our objective is to find both an $ϵ>0$ and an $x\in ℝ$, such that $|x-x_0|<\delta$ and $|f(x)-f(x_0)|\ge ϵ$ . Here, $x$ may be chosen depending on $\delta$ , while $ϵ$ has to be the same for all $\delta >0$ . For a solution, we may proceed as follows

Step 1: Simplify the target inequality

First, we may simplify the two inequalities which have to be fulfilled by plugging in $x_0=0$ and $f(x_0)=0$. We therefore get: $|x|<\delta$ and $|f(x)|\ge ϵ$.

Step 2: Choose a suitable $ϵ>0$

We consider the graph of the function $f$ . It will help us finding the building bricks for our proof:

Graph of the topological sine function

We need to find an $ϵ>0$ , such that there are arguments in each arbitrarily narrow interval $(x_0-\delta ,x_0+\delta )=(-\delta ,\delta )$ whose function values have a distance larger than $ϵ$ from $f(x_0)=f(0)=0$ . Visually, no matter how small the width $\delta$ of the $2ϵ$-$2\delta$-rectangle is chosen, there will always be some points below or above it.

Taking a look at the graph, we see that our function oscillates between $-1$ and $1$ . Hence, $ϵ\le 1$ may be useful. In that case, there will be function values with $|f(x)|\ge 1$ in every arbitrarily small neighborhood around $x_0=0$. We choose $ϵ=\frac{1}{2}$. This is visualized in the following figure:

The topological sine function oscillates between -1 and 1 nea the origin. Hence, in each neighborhood around 0, there will be arguments x , where f(x) has a distance larger than 1/2 to 0.

After $ϵ$ has been chosen, an arbitrary $\delta >0$ will be assumed, for which we need to find a suitable $x$. This is what we will do next.

Step 3: Choice of a suitable $x\in ℝ$

We just set $ϵ=\frac{1}{2}$ . Therefore, $|\sin \left(\frac{1}{x}\right)|\ge \frac{1}{2}$ has to hold. So it would be nice to choose an $x$ with $\sin \left(\frac{1}{x}\right)=1$ . Now, $\sin (a)=1$ is obtained for any $a=\frac{\pi }{2}+2k\pi$ with $k\in \mathbb{Z}$ . The condition for $x$ such that the function gets 1 is therefore:

$\frac{1}{x}=\frac{\pi }{2}+2k\pi ⟺x=\frac{1}{\frac{\pi }{2}+2k\pi }$

So we found several $x$ , where $|f(x)|\ge ϵ$ . Now we only need to select one among them, which satisfies $|x|<\delta$ for the given $\delta$ . Our $x$ depend on $k$ . So we have to select a suitable $k\in \mathbb{Z}$ , where $|x|<\delta$ . To do so, let us plug $x=\frac{1}{\frac{\pi }{2}+2k\pi }$ into this inequality and solve it for $k$ :

$\begin{array}{cc}\left|x\right|<\delta & ⟺\left|\frac{1}{\frac{\pi }{2}+2k\pi }\right|<\delta \\ & ⟺\frac{1}{\frac{\pi }{2}+2k\pi }<\delta \\ & ⟺2k\pi +\frac{\pi }{2}>\frac{1}{\delta }\\ & ⟺2k\pi >\frac{1}{\delta }-\frac{\pi }{2}\\ & ⟺k>\frac{1}{2\pi }(\frac{1}{\delta }-\frac{\pi }{2})\end{array}$

So the condition on $k$ is $k>\frac{1}{2\pi }(\frac{1}{\delta }-\frac{\pi }{2})$. If we choose just any natural number $k$ above this threshold $\frac{1}{2\pi }(\frac{1}{\delta }-\frac{\pi }{2})$ , then $|x|<\delta$ will be fulfilled. Such a $k$ has to exist by Archimedes' axiom (for instance by flooring up the right-hand expression). So let us choose such a $k$ and define $x$ via $x=\frac{1}{\frac{\pi }{2}+2k\pi }$ . This gives us both $|x|<\delta$ and $|f(x)|\ge ϵ$ . So we got all building bricks together, which we will now assemble to a final proof:

Proof (Discontinuity of the topological sine fucntion)

Choose $ϵ=\frac{1}{2}$ and let $\delta >0$ be arbitrary. Choose a natural number $k$ with $k>\frac{1}{2\pi }(\frac{1}{\delta }-\frac{\pi }{2})$. Such a natural number $k$ has to exist by Archimedes' axiom. Further, let $x=\frac{1}{\frac{\pi }{2}+2k\pi }$. Then:

$\begin{array}{cc}k>\frac{1}{2\pi }(\frac{1}{\delta }-\frac{\pi }{2})& ⟹2k\pi >\frac{1}{\delta }-\frac{\pi }{2}\\ & ⟹2k\pi +\frac{\pi }{2}>\frac{1}{\delta }\\ & \downarrow \frac{1}{a}>\frac{1}{b}>0⟺0<a<b\\ & ⟹\frac{1}{\frac{\pi }{2}+2k\pi }<\delta \\ & \downarrow \frac{1}{\frac{\pi }{2}+2k\pi }>0\\ & ⟹\left|\frac{1}{\frac{\pi }{2}+2k\pi }\right|<\delta \\ & \downarrow x=\frac{1}{\frac{\pi }{2}+2k\pi }\\ & ⟹\left|x\right|<\delta \end{array}$

In addition:

$\begin{array}{cc}|f(x)-f(0)|& =|\sin \left(\frac{1}{x}\right)-0|\\ & \downarrow x=\frac{1}{\frac{\pi }{2}+2k\pi }\\ & =|\sin (\frac{\pi }{2}+2k\pi )|\\ & \downarrow \sin (\frac{\pi }{2}+2k\pi )=1\\ & =\left|1\right|\ge \frac{1}{2}=ϵ\end{array}$

Hence, the function is discontinuous at $x_0=0$.

Epsilon-Delta Criterion: Hyperbola

Exercise (Continuity of the hyperbolic function)

Prove that the function $f:{ℝ}^{+}\to ℝ$ with $f(x)=\frac{1}{x}$ is continuous.

How to get to the proof? (Continuity of the hyperbolic function)

The basic pattern for epsilon-delta proofs is applied here, as well. We would like to show the implication $|x-x_0|<\delta ⟹|f(x)-f(x_0)|<ϵ$ . Forst, let us plug in what we know and reshape our terms a bit until a $|x-x_0|$ appears:

$\begin{array}{cc}|f(x)-f(x_0)|& =|\frac{1}{x}-\frac{1}{x_0}|\\ & =\left|\frac{x-x_0}{x\cdot x_0}\right|\\ & =\frac{|x-x_0|}{|x||x_0|}\\ & =\frac{1}{|x||x_0|}\cdot |x-x_0|\end{array}$

By assumption, there will be $|x-x_0|<\delta$, of which we can make use:

$|f(x)-f(x_0)|=\dots =\frac{1}{|x||x_0|}\cdot |x-x_0|<\frac{1}{|x||x_0|}\cdot \delta$

The choice of $\delta$ may again only depend on $ϵ$ and $x_0$ so we need a smart estimate for $\frac{1}{|x|}$ in order to get rid of the $x$-dependence. To do so, we consider $\delta \le \frac{x_0}{2}$.

Why was $\delta \le \frac{x_0}{2}$ and not $\delta \le 1$ chosen? The explanation is quite simple: We need a $\delta$-neighborhood inside the domain of definition of $f$. If we had simply chosen $\delta \le 1$ , we might have get kicked out of this domain. For instance, in case$x_0=\frac{1}{2}$, the following problem appears:

The biggest $x$-value with $|x-\frac{1}{2}|<1$ is $x_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{3}{2}$ and the smallest one is $x_{\mathrm{m}\mathrm{i}\mathrm{n}}=-\frac{1}{2}$. However, $x_{\mathrm{m}\mathrm{i}\mathrm{n}}$ is not inside the domain of definition as $f:{ℝ}^{+}\to ℝ$. In particular, $x=0$ is element of that interval, where $f(x)=\frac{1}{x}$ cannot be defined at all.

A smarter choice for $\delta$, such that the $\delta$-neighborhood doesn't touch the $y$-axis is half of the distance to it, i.e. $\delta \le \frac{x_0}{2}$. A third of this distance or other fractions smaller than 1 would also be possible: $\delta \le \frac{x_0}{3}$, $\delta <\frac{x_0}{20}$ or $\delta <\frac{x_0}{5321}$.

As we chose $\delta \le \frac{x_0}{2}$ and by $|x-x_0|<\delta$ , there is $x\in [\frac{x_0}{2};\frac{3x_0}{2}]$. This allows for an upper bound: $\frac{1}{|x|}\le \frac{2}{|x_0|}$ and we may write:

$\begin{array}{cc}|f(x)-f(x_0)|& =\dots <\frac{1}{|x||x_0|}\cdot \delta \\ & \downarrow \delta =\frac{ϵ\cdot |x_0{|}^2}{2}\\ & \le \frac{2}{|x_0{|}^2}\cdot \delta \end{array}$

So we get the estimate:

$|f(x)-f(x_0)|\le \frac{2}{|x_0{|}^2}\cdot \delta$

Now, we want to prove $|f(x)-f(x_0)|<ϵ$ . Hence we choose $\delta \le \frac{ϵ\cdot |x_0{|}^2}{2}$. Plugging this in, our final inequality $|f(x)-f(x_0)|<ϵ$ will be fulfilled.

So again, we imposed two conditions for $\delta$ : $\delta \le \frac{x_0}{2}$ and $\delta \le \frac{ϵ\cdot |x_0{|}^2}{2}$. Both are fulfilled by $\delta =min\{\frac{x_0}{2},\frac{ϵ\cdot |x_0{|}^2}{2}\}$, which we will use in the proof:

Proof (Continuity of the hyperbolic function)

Let $f:{ℝ}^{+}\to ℝ$ with $f(x)=\frac{1}{x}$ and let $x_0\in {ℝ}^{+}$ be arbitrary. Further, let $ϵ>0$ be arbitrary. We choose $\delta :=min\{\frac{x_0}{2},\frac{ϵ\cdot |x_0{|}^2}{2}\}$. For all $x\in ℝ$ with $|x-x_0|<\delta$ there is:

$\begin{array}{cc}|f(x)-f(x_0)|& =|\frac{1}{x}-\frac{1}{x_0}|\\ & =\left|\frac{x-x_0}{x\cdot x_0}\right|\\ & =\frac{|x-x_0|}{|x||x_0|}\\ & =\frac{1}{|x||x_0|}|x-x_0|\\ & \downarrow \frac{1}{|x|}\le \frac{2}{|x_0|}\\ & \le \frac{2}{|x_0{|}^2}|x-x_0|\\ & \downarrow |x-x_0|<\delta \\ & <\frac{2\delta }{|x_0{|}^2}\\ & \downarrow \delta =\frac{ϵ\cdot |x_0{|}^2}{2}\\ & =\frac{2}{|x_0{|}^2}\cdot \frac{ϵ\cdot |x_0{|}^2}{2}\\ & =ϵ\end{array}$

Hence, the function $f$ is continuous at $x_0$ . And as $x_0$ was chosen to be arbitrary, the whole function $f$ is continuous.

Epsilon-Delta Criterion: Concatenated Root Function

Exercise (Epsilon-Delta proof for the continuity of the Square Root Function)

Show, using the epsilon-delta criterion, that the following function is continuous:

$f:ℝ\to ℝ,x\mapsto \sqrt{5+x^2}$

How to get to the proof? (Epsilon-Delta proof for the continuity of the Square Root Function)

We need to show, that for any given $ϵ>0$ , there is a $\delta >0$ , such that all $x\in ℝ$ with $|x-a|<\delta$ satisfy the inequality $|f(x)-f(a)|<ϵ$ . So let us take a look at the target inequality $|f(x)-f(a)|<ϵ$ and estimate the absolute $|f(x)-f(a)|$ from above. We are able to control the term $|x-a|$ . Therefor, would like to get an upper bound for $|f(x)-f(a)|$ including the expression $|x-a|$ . So we are looking for an inequality of the form

$|f(x)-f(a)|\le K(x,a)\cdot |x-a|$

Here, $K(x,a)$ is some expression depending on $x$ and $a$ . The second factor is smaller than $\delta$ and can be made arbitrarily small by a suitable choice of $\delta$ . Such a bound is constructed as follows:

$\begin{array}{cc}|f(x)-f(a)|& =|\sqrt{5+x^2}-\sqrt{5+a^2}|\\ & \downarrow \text{expand with}|\sqrt{5+x^2}+\sqrt{5+a^2}|\\ & =\frac{|\sqrt{5+x^2}-\sqrt{5+a^2}|\cdot |\sqrt{5+x^2}+\sqrt{5+a^2}|}{|\sqrt{5+x^2}+\sqrt{5+a^2}|}\\ & \downarrow \sqrt{5+x^2}+\sqrt{5+a^2}\ge 0\\ & =\frac{|x^2-a^2|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & =\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\cdot |x-a|\\ & \downarrow K(x,a):=\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & =K(x,a)\cdot |x-a|\end{array}$

Since $|x-a|<\delta$ , there is:

$|f(x)-f(a)|=K(x,a)\cdot |x-a|<K(x,a)\cdot \delta$

If we now choose $\delta$ small enough, such that $K(x,a)\cdot \delta \le ϵ$ , then we obtain our target inequality $|f(x)-f(a)|\le ϵ$. But $K(x,a)$ still depends on $x$ , so $\delta$ would have to depend on , too - and we required one choice of $\delta$ which is suitable for all $x$ . Therefore, we need to get rid of the $x$-dependence. This is done by an estimate of the first factor, such that our inequality takes the form $K(x,a)\le \tilde{K}(a)$ :

$\begin{array}{cc}K(x,a)& =\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & \downarrow \text{triangle inequality}\\ & \le \frac{\left|x\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}}+\frac{\left|a\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & \downarrow \frac{a}{b+c}\le \frac{a}{b}\text{ for }a,c\ge 0,b>0\\ & \le \frac{\left|x\right|}{\sqrt{5+x^2}}+\frac{\left|a\right|}{\sqrt{5+a^2}}\\ & \downarrow \frac{|a|}{\sqrt{5+a^2}}\le 1\text{, since }\sqrt{5+a^2}\ge \sqrt{a^2}=|a|\\ & \le 2=:\tilde{K}(a)\end{array}$

We even made $\tilde{K}(a)$ independent of $a$ , which would in fact not have been necessary. So we obtain the following inequality

$|f(x)-f(a)|\le 2\cdot |x-a|<2\cdot \delta$

We need the estimate $2\cdot \delta \le ϵ$, in order to fulfill the target inequality $|f(x)-f(a)|<ϵ$ . The choice of $\delta =\frac{ϵ}{2}$ is sufficient for that. So let us write down the proof:

Proof (Epsilon-Delta proof for the continuity of the Square Root Function)

Let $f:ℝ\to ℝ$ with $f(x)=\sqrt{5+x^2}$. Let $a\in ℝ$ and an arbitrary $ϵ>0$ be given. We choose $\delta =\frac{ϵ}{2}$. For all $x\in ℝ$ with $|x-a|<\delta$ there is:

$\begin{array}{cc}|f(x)-f(a)|& =|\sqrt{5+x^2}-\sqrt{5+a^2}|\\ & =\frac{|\sqrt{5+x^2}-\sqrt{5+a^2}|\cdot |\sqrt{5+x^2}+\sqrt{5+a^2}|}{|\sqrt{5+x^2}+\sqrt{5+a^2}|}\\ & =\frac{|x^2-a^2|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\\ & =\frac{|x+a|}{\sqrt{5+x^2}+\sqrt{5+a^2}}\cdot |x-a|\\ & \le (\frac{\left|x\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}}+\frac{\left|a\right|}{\sqrt{5+x^2}+\sqrt{5+a^2}})\cdot |x-a|\\ & \le (\frac{\left|x\right|}{\sqrt{5+x^2}}+\frac{\left|a\right|}{\sqrt{5+a^2}})\cdot |x-a|\\ & \le (1+1)\cdot |x-a|\\ & \le 2\cdot |x-a|\\ & \downarrow |x-a|<\delta =\frac{ϵ}{2}\\ & <2\cdot \frac{ϵ}{2}=ϵ\end{array}$

Hence, $f$ is a continuous function.