Proving continuity – Serlo

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 $\lim _{n\to \infty }f(x_{n})=f(x_{0})=f\left(\lim _{n\to \infty }x_{n}\right)$.
• Showing the epsilon-delta criterion: For every $\epsilon >0$ 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 $|f(x)-f(x_{0})|<\epsilon$ is satisfied.

Composition of continuous functions

→ Main article: Composition of continuous functions

General proof sketch

Following the concatenation theorems, every composition of continuous functions is again continuous function. So if $f:D\to \mathbb {R}$ 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 $f:\ldots$ with $f(x)=\ldots$. The function $f$ is a concatenation of the following functions:

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

Since $f(x)=\ldots$ (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:

$f:\mathbb {R} \to \mathbb {R} ,x\mapsto {\sqrt {5+x^{2}}}$

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 $\epsilon >0$ , there is a $\delta >0$ , such that all $x\in \mathbb {R}$ with $|x-a|<\delta$ satisfy the inequality $|f(x)-f(a)|<\epsilon$ . So let us take a look at the target inequality $|f(x)-f(a)|<\epsilon$ 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

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

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:

${\begin{aligned}|f(x)-f(a)|&=\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\text{expand with}}\left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|\right.}\\[0.3em]&={\frac {\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\cdot \left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}{\left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\geq 0\right.}\\[0.3em]&={\frac {\left|x^{2}-a^{2}\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\cdot |x-a|\\[0.3em]&\quad {\color {Gray}\left\downarrow \ K(x,a):={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\right.}\\[0.3em]&=K(x,a)\cdot |x-a|\end{aligned}}$

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 $K(x,a)\cdot \delta \leq \epsilon$ , then we obtain our target inequality $|f(x)-f(a)|\leq \epsilon$. 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 $K(x,a)\leq {\tilde {K}}(a)$ :

${\begin{aligned}K(x,a)&={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\text{triangle inequality}}\right.}\\[0.3em]&\leq {\frac {\left|x\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}+{\frac {\left|a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\frac {a}{b+c}}\leq {\frac {a}{b}}{\text{ for }}a,c\geq 0,b>0\right.}\\[0.3em]&\leq {\frac {\left|x\right|}{\sqrt {5+x^{2}}}}+{\frac {\left|a\right|}{\sqrt {5+a^{2}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\frac {|a|}{\sqrt {5+a^{2}}}}\leq 1{\text{, since }}{\sqrt {5+a^{2}}}\geq {\sqrt {a^{2}}}=|a|\right.}\\[0.3em]&\leq 2=:{\tilde {K}}(a)\end{aligned}}$

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

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

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

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

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

${\begin{aligned}|f(x)-f(a)|&=\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\\[0.3em]&={\frac {\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\cdot \left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}{\left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}}\\[0.3em]&={\frac {\left|x^{2}-a^{2}\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\cdot |x-a|\\[0.3em]&\leq \left({\frac {\left|x\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}+{\frac {\left|a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\right)\cdot |x-a|\\[0.3em]&\leq \left({\frac {\left|x\right|}{\sqrt {5+x^{2}}}}+{\frac {\left|a\right|}{\sqrt {5+a^{2}}}}\right)\cdot |x-a|\\[0.3em]&\leq (1+1)\cdot |x-a|\\[0.3em]&\leq 2\cdot |x-a|\\[0.3em]&\quad {\color {Gray}\left\downarrow \ |x-a|<\delta ={\frac {\epsilon }{2}}\right.}\\[0.3em]&<2\cdot {\frac {\epsilon }{2}}=\epsilon \end{aligned}}$

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 $f:\mathbb {R} \setminus \{0\}\to \mathbb {R}$ 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:

"For every $x_{0}\in \mathbb {R} \setminus \{0\}$ there exists a neighborhood around $x_{0}$ where $f$ is either constant $1$ or constant $-1$. Since constant functions are continuous, $f$ must also be continuous at the point $x_{0}$. Therefore,$f$ is continuous“.

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:

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

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 $g$:

$g:\mathbb {R} \to \mathbb {R} :x\mapsto {\begin{cases}x^{2}&x\geq 0\\|x|&x<0\end{cases}}$

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

→ Main article: Sequential definition of continuity

Review: Sequence Criterion

Definition (Sequence criterion of continuity)

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

$\lim _{n\to \infty }f(x_{n})=f\left(\lim _{n\to \infty }x_{n}\right)=f(x)$

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 $\lim _{n\to \infty }x_{n}=x_{0}$ , there is $\lim _{n\to \infty }f(x_{n})=f(x_{0})$ . A proof for this could schematically look as follows:

Let $f:\ldots$ be a function defined by $f(x)=\ldots$ and $x_{0}=\ldots$ . In addition, let $(x_{n})_{n\in \mathbb {N} }$ be any sequence of arguments satisfying $\lim _{n\to \infty }x_{n}=x_{0}$. Then, there is:

$\lim _{n\to \infty }f(x_{n})=\ldots =f(x_{0})$

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 $f:\ldots$ be a function defined by $f(x)=\ldots$ 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 $\lim _{n\to \infty }x_{n}=x_{0}$. Then, there is:

$\lim _{n\to \infty }f(x_{n})=\ldots =f(x_{0})$

Example Problem

Exercise (Continuity of quadratic functions)

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

Proof (Continuity of quadratic functions)

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

${\begin{aligned}\lim _{n\to \infty }f(x_{n})&=\lim _{n\to \infty }x_{n}^{2}\\&=\lim _{n\to \infty }x_{n}\cdot x_{n}\\[0.5em]&{\color {OliveGreen}\left\downarrow \ \lim _{n\to \infty }a_{n}\cdot b_{n}=\lim _{n\to \infty }a_{n}\cdot \lim _{n\to \infty }b_{n}\right.}\\[0.5em]&=\left(\lim _{n\to \infty }x_{n}\right)\cdot \left(\lim _{n\to \infty }x_{n}\right)\\&{\color {OliveGreen}\left\downarrow \ {\text{assumption}}\lim _{n\to \infty }x_{n}=x\right.}\\[0.5em]&=x\cdot x=x^{2}=f(x)\end{aligned}}$

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

Epsilon-Delta Criterion

→ Main article: Epsilon-delta definition of continuity

Review: Epsilon-Delta Criterion

Definition (Epsilon-Delta-definition of continuity)

A function $f:D\to \mathbb {R}$ with $D\subseteq \mathbb {R}$ is continuous at $x_{0}\in D$, if and only if for any $\epsilon >0$ there is a $\delta >0$ , such that $|f(x)-f(x_{0})|<\epsilon$ 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

$\forall \epsilon >0\,\exists \delta >0\,\forall x\in D:|x-x_{0}|<\delta \implies |f(x)-f(x_{0})|<\epsilon$

General Proof Structure

In mathematical quantifiers, the epsilon-delta definition of continuity of a function $f$ at the point $x_{0}$ reads:

${\color {Red}\forall \epsilon >0}\ {\color {RedOrange}\exists \delta >0}\ {\color {OliveGreen}\forall x\in D:|x-x_{0}|<\delta }{\color {Blue}{}\implies |f(x)-f(x_{0})|<\epsilon }$

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

${\begin{array}{l}{\color {Red}\underbrace {{\underset {}{}}{\text{Let }}\epsilon >0{\text{ be arbitrary.}}} _{\forall \epsilon >0}}\ {\color {RedOrange}\underbrace {{\underset {}{}}{\text{Choose }}\delta =\ldots {\text{ There is }}\delta {\text{, since}}\ldots } _{\exists \delta >0}}\\{\color {OliveGreen}\underbrace {{\underset {}{}}{\text{Let }}x\in D{\text{ with }}|x-x_{0}|<\delta {\text{ be arbitrary.}}} _{\forall x\in D:|x-x_{0}|<\delta }}\ {\color {Blue}\underbrace {{\underset {}{}}{\text{There is: }}|f(x)-f(x_{0})|<\ldots <\epsilon } _{{}\implies |f(x)-f(x_{0})|<\epsilon }}\end{array}}$

Example Problem and General Procedure

Exercise (Continuity of the quadratic function)

Prove that the function $f:\mathbb {R} \to \mathbb {R}$ 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 $x_{0}\in \mathbb {R}$ . Using the proof structure for the epsilon-delta criterion, we are given an arbitrary $\epsilon >0$ . Our job is to find a suitable $\delta >0$ , such that the inequality $|f(x)-f(x_{0})|<\epsilon$ 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 $\epsilon$:

$|f(x)-f(x_{0})|=\left|x^{2}-x_{0}^{2}\right|$

The expression $|x-x_{0}|$ may easily be controlled by $\delta$. Hence, it makes sense to construct an upper estimate for $\left|x^{2}-x_{0}^{2}\right|$ 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 $\epsilon$ 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:

$|x+x_{0}|\cdot \delta =|x\underbrace {-x_{0}+x_{0}} _{=\ 0}+x_{0}|\cdot \delta =|x-x_{0}+2x_{0}|\cdot \delta$

The absolute $|x-x_{0}|$ is obtained using the triangle inequality. This absolute $|x-x_{0}|$ is again bounded from above by $\delta$ :

$|x-x_{0}+2x_{0}|\cdot \delta \leq (|x-x_{0}|+|2x_{0}|)\cdot \delta <(\delta +2|x_{0}|)\cdot \delta$

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 $(\delta +2|x_{0}|)\cdot \delta \leq \epsilon$ , we will get the final inequality $|f(x)-f(x_{0})|<\epsilon$ . This $\delta$ is practically found solving the quadratic equation $\delta ^{2}+2|x_{0}|\delta =\epsilon$ 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 $\delta \leq 1$, then $\delta +2|x_{0}|\leq 1+2|x_{0}|$ which simplifies things:

$|f(x)-f(x_{0})|<(\underbrace {\delta +2|x_{0}|} _{\leq 1+2|x_{0}|})\cdot \delta \leq (1+2|x_{0}|)\cdot \delta$

So $(1+2|x_{0}|)\cdot \delta \leq \epsilon$ will also do the job. This inequality can be solved for $\delta$ to get the second condition on $\delta$(the first one was $\delta \leq 1$):

$(1+2|x_{0}|)\cdot \delta \leq \epsilon \iff \delta \leq {\frac {\epsilon }{1+2|x_{0}|}}$

So any $\delta$ fulfilling both conditions does the job: $\delta \leq 1$ and $\delta ={\tfrac {\epsilon }{1+2|x_{0}|}}$ have to hold. Ind indeed, both are true for $\delta :=\min \left\{1,{\tfrac {\epsilon }{1+2|x_{0}|}}\right\}$. This choice will be included into the final proof:

Proof (Continuity of the quadratic function)

Let $\epsilon >0$ be arbitrary and $\delta :=\min \left\{1,{\tfrac {\epsilon }{1+2|x_{0}|}}\right\}$. If an argument $x$ fulfills $\left|x-x_{0}\right|<\delta$ then:

${\begin{aligned}|f(x)-f(x_{0})|&=|x^{2}-x_{0}^{2}|\\[0.5em]&=|x+x_{0}||x-x_{0}|\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ |x-x_{0}|<\delta \right.}\\[0.5em]&<|x+x_{0}|\cdot \delta \\[0.5em]&=|x\underbrace {-x_{0}+x_{0}} _{=\ 0}+x_{0}|\cdot \delta \\[0.5em]&=|x-x_{0}+2x_{0}|\cdot \delta \\[0.5em]&\leq (|x-x_{0}|+2|x_{0}|)\cdot \delta \\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ |x-x_{0}|<\delta \right.}\\[0.5em]&<(\delta +2|x_{0}|)\cdot \delta \\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ \delta \leq 1\right.}\\[0.5em]&\leq (1+2|x_{0}|)\cdot \delta \\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ \delta ={\frac {\epsilon }{1+2|x_{0}|}}\right.}\\[0.5em]&=(1+2|x_{0}|)\cdot {\frac {\epsilon }{1+2|x_{0}|}}\\[0.5em]&=\epsilon \end{aligned}}$

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 $x_{0}=1$:

$f:\mathbb {R} \to \mathbb {R} :x\mapsto f(x)=5\left|x^{2}-2\right|+3$

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

We need to show that for each given $\epsilon >0$ , there is a $\delta >0$ , such that for all $x\in \mathbb {R}$ with $|x-x_{0}|<\delta$ the inequality $|f(x)-f(x_{0})|<\epsilon$ 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 $|f(x)-f(x_{0})|<\epsilon$ :

${\begin{aligned}|f(x)-f(x_{0})|&=|f(x)-f(1)|\\[0.5em]&=|5\cdot |x^{2}-2|+3-(5|1^{2}-2|+3)|\\[0.5em]&=|5\cdot |x^{2}-2|-5|\\[0.5em]&=5\cdot ||x^{2}-2|-1|\end{aligned}}$

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 $||a|-|b||\leq |a-b|$ . For instance, we could use the following estimate:

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots \\[0.3em]&=5\cdot ||x^{2}-2|-1|\\[0.3em]&=5\cdot ||x^{2}-2|-|1||\\[0.3em]&\ {\color {Gray}\left\downarrow \ ||a|-|b||\leq |a-b|\right.}\\[0.3em]&\leq 5\cdot |x^{2}-2-1|\\[0.3em]&=5\cdot |x^{2}-3|\end{aligned}}$

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 $||a|-|b||\leq |a-b|$ :

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots \\[0.3em]&=5\cdot |x^{2}-2|-1|\\[0.3em]&=5\cdot |x^{2}-2|-|-1||\\[0.3em]&\quad {\color {Gray}\left\downarrow \ |a-b|\geq ||a|-|b||\right.}\\[0.3em]&\leq 5\cdot |x^{2}-2-(-1)|\\[0.3em]&=5\cdot |x^{2}-1|\end{aligned}}$

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

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots \\[0.5em]&\leq 5\cdot |x^{2}-1|\\[0.5em]&=5\cdot |x+1||x-1|\end{aligned}}$

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

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots \\[0.5em]&\leq 5\cdot |x^{2}-1|\\[0.5em]&=5\cdot |x+1||x-1|\\[0.5em]&<5\cdot |x+1|\cdot \delta \end{aligned}}$

Now, the required $\delta$ must only depend on $\epsilon$ 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, $\delta \leq 1$ turns out to be quite useful. In fact, $\delta \leq 2$ 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 $\delta \leq 1$ , we now have $|x-1|<1$ and as $x\in [1-1;1+1]$ , we obtain $1+x\in [1;3]$ and $|x+1|\leq 3$. This is the upper bound we were looking for:

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots \\[0.5em]&<5\cdot |x+1|\cdot \delta \\[0.5em]&\quad {\color {Gray}\left\downarrow \ |x+1|\leq 3\right.}\\[0.5em]&\leq 15\cdot \delta \end{aligned}}$

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

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots \\[0.5em]&<15\cdot \delta \\[0.5em]&\leq 15\cdot {\frac {\epsilon }{15}}\\[0.5em]&\leq \epsilon .\end{aligned}}$

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

Proof (Example for a proof of continuity)

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

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

As $\delta =\min \left\{1,{\tfrac {\epsilon }{15}}\right\}$ , there is $\delta \leq 1$. Hence $|x-1|\leq 1$ and $x\in [1-1;1+1]$. It follows that $1+x\in [1;3]$ and therefore $|x+1|\leq 3$.

Step 2: $|f(x)-f(x_{0})|<\epsilon$

${\begin{aligned}|f(x)-f(x_{0})|&=|f(x)-f(1)|\\[0.5em]&=|5\cdot |x^{2}-2|+3-(5|1^{2}-2|+3)|\\[0.5em]&=|5\cdot |x^{2}-2|-5|\\[0.5em]&=5\cdot ||x^{2}-2|-1|\\[0.5em]&\quad {\color {Gray}\left\downarrow \ 1\iff |-1|\right.}\\[0.5em]&=5\cdot ||x^{2}-2|-|-1||\\[0.5em]&\quad {\color {Gray}\left\downarrow \ |a-b|\geq ||a|-|b||\right.}\\[0.5em]&\leq 5\cdot |x^{2}-2-(-1)|\\[0.5em]&=5\cdot |x^{2}-1|\\[0.5em]&\quad {\color {Gray}\left\downarrow \ a^{2}-b^{2}=(a+b)(a-b)\right.}\\[0.5em]&=5\cdot (|x+1|\cdot \underbrace {|x-1|} _{<\delta })\\[0.5em]&<5\cdot |x+1|\cdot \delta \\[0.5em]&\quad {\color {Gray}\left\downarrow \ |x+1|\leq 3\right.}\\[0.5em]&\leq 15\delta \\[0.5em]&\quad {\color {Gray}\left\downarrow \ \delta ={\frac {\epsilon }{15}}\right.}\\[0.5em]&=15\cdot {\frac {\epsilon }{15}}\\[0.5em]&\leq \epsilon \end{aligned}}$

Hence, the function is continuous at $x_{0}=1$ .

Hyperbola

Exercise (Continuity of the hyperbolic function)

Prove that the function $f:\mathbb {R} ^{+}\to \mathbb {R}$ with $f(x)={\tfrac {1}{x}}$ 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 $|x-x_{0}|<\delta \implies |f(x)-f(x_{0})|<\epsilon$ . Forst, let us plug in what we know and reshape our terms a bit until a $|x-x_{0}|$ appears:

${\begin{aligned}|f(x)-f(x_{0})|&=\left|{\frac {1}{x}}-{\frac {1}{x_{0}}}\right|\\[0.5em]&=\left|{\frac {x-x_{0}}{x\cdot x_{0}}}\right|\\[0.5em]&={\frac {|x-x_{0}|}{|x||x_{0}|}}\\[0.5em]&={\frac {1}{|x||x_{0}|}}\cdot |x-x_{0}|\end{aligned}}$

By assumption, there will be $|x-x_{0}|<\delta$, of which we can make use:

$|f(x)-f(x_{0})|=\ldots ={\frac {1}{|x||x_{0}|}}\cdot |x-x_{0}|<{\frac {1}{|x||x_{0}|}}\cdot \delta$

The choice of $\delta$ may again only depend on $\epsilon$ and $x_{0}$ so we need a smart estimate for ${\tfrac {1}{|x|}}$ in order to get rid of the $x$-dependence. To do so, we consider $\delta \leq {\tfrac {x_{0}}{2}}$.

Why was $\delta \leq {\tfrac {x_{0}}{2}}$ and not $\delta \leq 1$ chosen? The explanation is quite simple: We need a $\delta$-neighborhood inside the domain of definition of $f$. If we had simply chosen $\delta \leq 1$ , we might have get kicked out of this domain. For instance, in case$x_{0}={\tfrac {1}{2}}$, the following problem appears:

The biggest $x$-value with $\left|x-{\tfrac {1}{2}}\right|<1$ is $x_{\mathrm {max} }={\tfrac {3}{2}}$ and the smallest one is $x_{\mathrm {min} }=-{\tfrac {1}{2}}$. However, $x_{\mathrm {min} }$ is not inside the domain of definition as $f:\mathbb {R} ^{+}\to \mathbb {R}$. In particular, $x=0$ is element of that interval, where $f(x)={\tfrac {1}{x}}$ 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. $\delta \leq {\tfrac {x_{0}}{2}}$. A third of this distance or other fractions smaller than 1 would also be possible: $\delta \leq {\tfrac {x_{0}}{3}}$, $\delta <{\tfrac {x_{0}}{20}}$ or $\delta <{\tfrac {x_{0}}{5321}}$.

As we chose $\delta \leq {\tfrac {x_{0}}{2}}$ and by $|x-x_{0}|<\delta$ , there is $x\in \left[{\tfrac {x_{0}}{2}};{\tfrac {3x_{0}}{2}}\right]$. This allows for an upper bound: ${\tfrac {1}{|x|}}\leq {\tfrac {2}{|x_{0}|}}$ and we may write:

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots <{\frac {1}{|x||x_{0}|}}\cdot \delta \\[0.5em]&\quad {\color {Gray}\left\downarrow \ \delta ={\frac {\epsilon \cdot |x_{0}|^{2}}{2}}\right.}\\[0.5em]&\leq {\frac {2}{|x_{0}|^{2}}}\cdot \delta \end{aligned}}$

So we get the estimate:

$|f(x)-f(x_{0})|\leq {\frac {2}{|x_{0}|^{2}}}\cdot \delta$

Now, we want to prove $|f(x)-f(x_{0})|<\epsilon$ . Hence we choose $\delta \leq {\tfrac {\epsilon \cdot |x_{0}|^{2}}{2}}$. Plugging this in, our final inequality $|f(x)-f(x_{0})|<\epsilon$ will be fulfilled.

So again, we imposed two conditions for $\delta$ : $\delta \leq {\tfrac {x_{0}}{2}}$ and $\delta \leq {\tfrac {\epsilon \cdot |x_{0}|^{2}}{2}}$. Both are fulfilled by $\delta =\min \left\{{\tfrac {x_{0}}{2}},{\tfrac {\epsilon \cdot |x_{0}|^{2}}{2}}\right\}$, which we will use in the proof:

Proof (Continuity of the hyperbolic function)

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

${\begin{aligned}|f(x)-f(x_{0})|&=\left|{\frac {1}{x}}-{\frac {1}{x_{0}}}\right|\\[0.5em]&=\left|{\frac {x-x_{0}}{x\cdot x_{0}}}\right|\\[0.5em]&={\frac {|x-x_{0}|}{|x||x_{0}|}}\\[0.5em]&={\frac {1}{|x||x_{0}|}}|x-x_{0}|\\[0.5em]&\quad {\color {Gray}\left\downarrow \ {\frac {1}{|x|}}\leq {\frac {2}{|x_{0}|}}\right.}\\[0.5em]&\leq {\frac {2}{|x_{0}|^{2}}}|x-x_{0}|\\[0.5em]&\quad {\color {Gray}\left\downarrow \ |x-x_{0}|<\delta \right.}\\[0.5em]&<{\frac {2\delta }{|x_{0}|^{2}}}\\[0.5em]&\quad {\color {Gray}\left\downarrow \ \delta ={\frac {\epsilon \cdot |x_{0}|^{2}}{2}}\right.}\\[0.5em]&={\frac {2}{|x_{0}|^{2}}}\cdot {\frac {\epsilon \cdot |x_{0}|^{2}}{2}}\\[0.5em]&=\epsilon \end{aligned}}$

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:

$f:\mathbb {R} \to \mathbb {R} ,x\mapsto {\sqrt {5+x^{2}}}$

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 $\epsilon >0$ , there is a $\delta >0$ , such that all $x\in \mathbb {R}$ with $|x-a|<\delta$ satisfy the inequality $|f(x)-f(a)|<\epsilon$ . So let us take a look at the target inequality $|f(x)-f(a)|<\epsilon$ 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

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

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:

${\begin{aligned}|f(x)-f(a)|&=\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\text{expand with}}\left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|\right.}\\[0.3em]&={\frac {\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\cdot \left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}{\left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\geq 0\right.}\\[0.3em]&={\frac {\left|x^{2}-a^{2}\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\cdot |x-a|\\[0.3em]&\quad {\color {Gray}\left\downarrow \ K(x,a):={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\right.}\\[0.3em]&=K(x,a)\cdot |x-a|\end{aligned}}$

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 $K(x,a)\cdot \delta \leq \epsilon$ , then we obtain our target inequality $|f(x)-f(a)|\leq \epsilon$. 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 $K(x,a)\leq {\tilde {K}}(a)$ :

${\begin{aligned}K(x,a)&={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\text{triangle inequality}}\right.}\\[0.3em]&\leq {\frac {\left|x\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}+{\frac {\left|a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\frac {a}{b+c}}\leq {\frac {a}{b}}{\text{ for }}a,c\geq 0,b>0\right.}\\[0.3em]&\leq {\frac {\left|x\right|}{\sqrt {5+x^{2}}}}+{\frac {\left|a\right|}{\sqrt {5+a^{2}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\frac {|a|}{\sqrt {5+a^{2}}}}\leq 1{\text{, since }}{\sqrt {5+a^{2}}}\geq {\sqrt {a^{2}}}=|a|\right.}\\[0.3em]&\leq 2=:{\tilde {K}}(a)\end{aligned}}$

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

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

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

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

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

${\begin{aligned}|f(x)-f(a)|&=\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\\[0.3em]&={\frac {\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\cdot \left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}{\left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}}\\[0.3em]&={\frac {\left|x^{2}-a^{2}\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\cdot |x-a|\\[0.3em]&\leq \left({\frac {\left|x\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}+{\frac {\left|a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\right)\cdot |x-a|\\[0.3em]&\leq \left({\frac {\left|x\right|}{\sqrt {5+x^{2}}}}+{\frac {\left|a\right|}{\sqrt {5+a^{2}}}}\right)\cdot |x-a|\\[0.3em]&\leq (1+1)\cdot |x-a|\\[0.3em]&\leq 2\cdot |x-a|\\[0.3em]&\quad {\color {Gray}\left\downarrow \ |x-a|<\delta ={\frac {\epsilon }{2}}\right.}\\[0.3em]&<2\cdot {\frac {\epsilon }{2}}=\epsilon \end{aligned}}$

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:

$f:\mathbb {R} \to \mathbb {R} :x\mapsto {\begin{cases}\sin \left({\frac {1}{x}}\right)&x\neq 0\\0&x=0\end{cases}}$

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 $\epsilon >0$ and an $x\in \mathbb {R}$, such that $|x-x_{0}|<\delta$ and $|f(x)-f(x_{0})|\geq \epsilon$ . Here, $x$ may be chosen depending on $\delta$ , while $\epsilon$ 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 $|f(x)|\geq \epsilon$.

Step 2: Choose a suitable $\epsilon >0$

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 $\epsilon >0$ , 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 $\epsilon$ from $f(x_{0})=f(0)=0$ . Visually, no matter how small the width $\delta$ of the $2\epsilon$-$2\delta$-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, $\epsilon \leq 1$ may be useful. In that case, there will be function values with $|f(x)|\geq 1$ in every arbitrarily small neighborhood around $x_{0}=0$. We choose $\epsilon ={\tfrac {1}{2}}$. 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 $\epsilon$ 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 $x\in \mathbb {R}$

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

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

So we found several $x$ , where $|f(x)|\geq \epsilon$ . 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 $x={\tfrac {1}{{\tfrac {\pi }{2}}+2k\pi }}$ into this inequality and solve it for $k$ :

${\begin{aligned}\left|x\right|<\delta &\iff \left|{\frac {1}{{\frac {\pi }{2}}+2k\pi }}\right|<\delta \\[0.5em]&\iff {\frac {1}{{\frac {\pi }{2}}+2k\pi }}<\delta \\[0.5em]&\iff 2k\pi +{\frac {\pi }{2}}>{\frac {1}{\delta }}\\[0.5em]&\iff 2k\pi >{\frac {1}{\delta }}-{\frac {\pi }{2}}\\[0.5em]&\iff k>{\frac {1}{2\pi }}\left({\frac {1}{\delta }}-{\frac {\pi }{2}}\right)\end{aligned}}$

So the condition on $k$ is $k>{\tfrac {1}{2\pi }}\left({\tfrac {1}{\delta }}-{\tfrac {\pi }{2}}\right)$. If we choose just any natural number $k$ above this threshold ${\tfrac {1}{2\pi }}\left({\tfrac {1}{\delta }}-{\tfrac {\pi }{2}}\right)$ , 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 $x={\tfrac {1}{{\tfrac {\pi }{2}}+2k\pi }}$ . This gives us both $|x|<\delta$ and $|f(x)|\geq \epsilon$ . So we got all building bricks together, which we will now assemble to a final proof:

Proof (Discontinuity of the topological sine fucntion)

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

${\begin{aligned}k>{\frac {1}{2\pi }}\left({\frac {1}{\delta }}-{\frac {\pi }{2}}\right)&\implies 2k\pi >{\frac {1}{\delta }}-{\frac {\pi }{2}}\\[0.5em]&\implies 2k\pi +{\frac {\pi }{2}}>{\frac {1}{\delta }}\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ {\frac {1}{a}}>{\frac {1}{b}}>0\iff 0<a<b\right.}\\[0.5em]&\implies {\frac {1}{{\frac {\pi }{2}}+2k\pi }}<\delta \\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ {\frac {1}{{\frac {\pi }{2}}+2k\pi }}>0\right.}\\[0.5em]&\implies \left|{\frac {1}{{\frac {\pi }{2}}+2k\pi }}\right|<\delta \\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ x={\frac {1}{{\frac {\pi }{2}}+2k\pi }}\right.}\\[0.5em]&\implies \left|x\right|<\delta \end{aligned}}$

In addition:

${\begin{aligned}\left|f(x)-f(0)\right|&=\left|\sin \left({\frac {1}{x}}\right)-0\right|\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ x={\frac {1}{{\frac {\pi }{2}}+2k\pi }}\right.}\\[0.5em]&=\left|\sin \left({\frac {\pi }{2}}+2k\pi \right)\right|\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ \sin \left({\frac {\pi }{2}}+2k\pi \right)=1\right.}\\[0.5em]&=\left|1\right|\geq {\frac {1}{2}}=\epsilon \end{aligned}}$

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 $f:\mathbb {R} \to \mathbb {R}$ with $f(x)=|x|$ be the absolute function. Let $x_{0}\in \mathbb {R}$ be a real number and $(x_{n})_{n\in \mathbb {N} }$ a sequence converging to it $\lim _{n\to \infty }x_{n}=x_{0}$. In chapter „Grenzwertsätze: Grenzwert von Folgen berechnen“ we have proven the absolute rule, stating that $\lim _{n\to \infty }|x_{n}|=|x_{0}|$ , whenever there is $\lim _{n\to \infty }x_{n}=x_{0}$ . Hence:

$\lim _{n\to \infty }f(x_{n})=\lim _{n\to \infty }|x_{n}|=|x_{0}|=f(x_{0})$

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:

$f:\mathbb {R} \to \mathbb {R} :x\mapsto {\begin{cases}\sin \left({\frac {1}{x}}\right)&x\neq 0\\0&x=0\end{cases}}$

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 $x\neq 0$ , $f$ is equal to the functionn $\sin \left({\tfrac {1}{x}}\right)$ in a sufficiently small neighbourhood of $x$. Since $\sin \left({\tfrac {1}{x}}\right)$ is just a composition of continuous functions, it is continuous itself and therefore, $f$ must be continuous for all $x\neq 0$ , 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 fincd a sequence of arguments $(x_{n})_{n\in \mathbb {N} }$ converging to $\lim _{n\to \infty }x_{n}=0$ but with $\lim _{n\to \infty }f(x_{n})\neq f(0)$ . 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 $\lim _{n\to \infty }f(x_{n})=1\neq 0=f(0)$ - and actually any other real number $f(x_{n})\neq 0$ 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 $x$ :

${\begin{aligned}{\begin{array}{rrrl}&&f(x)&=1\\[0.5em]{\overset {f(0)\neq 0}{\iff {}}}&&\sin \left({\frac {1}{x}}\right)&=1\\[0.5em]\iff {}&\exists k\in \mathbb {Z} :&{\frac {1}{x}}&={\frac {\pi }{2}}+2k\pi \\[0.5em]\iff {}&\exists k\in \mathbb {Z} :&x&={\frac {1}{{\frac {\pi }{2}}+2k\pi }}\end{array}}\end{aligned}}$

So for each $x={\tfrac {1}{{\frac {\pi }{2}}+2k\pi }}$ 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 $x_{n}={\tfrac {1}{{\frac {\pi }{2}}+2n\pi }}$. In that case:

$\lim _{n\to \infty }x_{n}=\lim _{n\to \infty }{\frac {1}{{\frac {\pi }{2}}+2n\pi }}=0$

And we have seen that $\lim _{n\to \infty }f(x_{n})=1\neq 0=f(0)$ . 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 $f:\mathbb {R} \to \mathbb {R}$ with $f(x)=\sin \left({\tfrac {1}{x}}\right)$ for $x\neq 0$ and $f(0)=0$. We consider the sequence $(x_{n})_{n\in \mathbb {N} }$ defined by $x_{n}={\tfrac {1}{{\frac {\pi }{2}}+2n\pi }}$. For this sequence:

$\lim _{n\to \infty }x_{n}=\lim _{n\to \infty }{\frac {1}{{\frac {\pi }{2}}+2n\pi }}=0$

And there is:

${\begin{aligned}\lim _{n\to \infty }f(x_{n})&=\lim _{n\to \infty }f\left({\frac {1}{{\frac {\pi }{2}}+2n\pi }}\right)\\[0.5em]&=\lim _{n\to \infty }\sin \left({\frac {1}{\frac {1}{{\frac {\pi }{2}}+2n\pi }}}\right)\\[0.5em]&=\lim _{n\to \infty }\sin \left({\frac {\pi }{2}}+2n\pi \right)\\[0.5em]&=\lim _{n\to \infty }1=1\end{aligned}}$

Hence, $\lim _{n\to \infty }f(x_{n})=1\neq 0=f(0)$ although $\lim _{n\to \infty }x_{n}=0$ . 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 $f:\mathbb {R} \to \mathbb {R}$ with $f(x)={\tfrac {1}{3}}x$ is continuous.

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

Graph of a function $f:\mathbb {R} \to \mathbb {R}$ with $f(x)={\tfrac {1}{3}}x$. 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 $x_{0}\in \mathbb {R}$ . So let $x_{0}$ be an arbitrary real number. Now, choose any arbitrary maximal error $\epsilon >0$ . Our task is now to find a sufficiently small $\delta >0$ , such that $|f(x)-f(x_{0})|<\epsilon$ for all arguments $x$ with $|x-x_{0}|<\delta$ . Let us take a closer look at the inequality $|f(x)-f(x_{0})|<\epsilon$ :

${\begin{aligned}{\begin{array}{rrl}&|f(x)-f(x_{0})|&<\epsilon \\[0.5em]\iff &\left|{\frac {1}{3}}x-{\frac {1}{3}}x_{0}\right|&<\epsilon \\[0.5em]\iff &\left|{\frac {1}{3}}(x-x_{0})\right|&<\epsilon \\[0.5em]\iff &{\frac {1}{3}}|x-x_{0}|&<\epsilon \end{array}}\end{aligned}}$

That means, ${\tfrac {1}{3}}|x-x_{0}|<\epsilon$ has to be fulfilled for all $x$ with $|x-x_{0}|<\delta$ . How to choose $\delta$ , such that $|x-x_{0}|<\delta$ implies ${\tfrac {1}{3}}|x-x_{0}|<\epsilon$ ?

We use that the inequality ${\tfrac {1}{3}}|x-x_{0}|<\epsilon$ 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 ${\tfrac {1}{3}}|x-x_{0}|$ :

${\frac {1}{3}}|x-x_{0}|{\stackrel {|x-x_{0}|<\delta }{<}}{\frac {1}{3}}\delta$

If $\delta$ is now chosen such that ${\tfrac {1}{3}}\delta \leq \epsilon$ , then ${\tfrac {1}{3}}|x-x_{0}|<{\tfrac {1}{3}}\delta$ will yield the inequality ${\tfrac {1}{3}}|x-x_{0}|<\epsilon$ which we wanted to show. The smallness condition for $\delta$ can now simply be found by resolving ${\tfrac {1}{3}}\delta \leq \epsilon$ for $\delta$:

${\tfrac {1}{3}}\delta \leq \epsilon \iff \delta \leq 3\epsilon$

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

Proof (Continuity of a linear function)

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

${\begin{aligned}|f(x)-f(x_{0})|&=\left|{\frac {1}{3}}x-{\frac {1}{3}}x_{0}\right|\\[0.5em]&=\left|{\frac {1}{3}}(x-x_{0})\right|\\[0.5em]&={\frac {1}{3}}|x-x_{0}|\\[0.5em]&\quad {\color {Gray}\left\downarrow \ |x-x_{0}|<\delta \right.}\\[0.5em]&<{\frac {1}{3}}\delta \\[0.5em]&\quad {\color {Gray}\left\downarrow \ \delta =3\epsilon \right.}\\[0.5em]&\leq {\frac {1}{3}}(3\epsilon )=\epsilon \end{aligned}}$

This shows $|f(x)-f(x_{0})|<\epsilon$, and establishes continuity of $f$ at $x_{0}$ by means of the epsilon-delta criterion. Since $x_{0}\in \mathbb {R}$ 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 $x_{0}=1$:

$f:\mathbb {R} \to \mathbb {R} :x\mapsto f(x)=5\left|x^{2}-2\right|+3$

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

We need to show that for each given $\epsilon >0$ , there is a $\delta >0$ , such that for all $x\in \mathbb {R}$ with $|x-x_{0}|<\delta$ the inequality $|f(x)-f(x_{0})|<\epsilon$ 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 $|f(x)-f(x_{0})|<\epsilon$ :

${\begin{aligned}|f(x)-f(x_{0})|&=|f(x)-f(1)|\\[0.5em]&=|5\cdot |x^{2}-2|+3-(5|1^{2}-2|+3)|\\[0.5em]&=|5\cdot |x^{2}-2|-5|\\[0.5em]&=5\cdot ||x^{2}-2|-1|\end{aligned}}$

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 $||a|-|b||\leq |a-b|$ . For instance, we could use the following estimate:

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots \\[0.3em]&=5\cdot ||x^{2}-2|-1|\\[0.3em]&=5\cdot ||x^{2}-2|-|1||\\[0.3em]&\ {\color {Gray}\left\downarrow \ ||a|-|b||\leq |a-b|\right.}\\[0.3em]&\leq 5\cdot |x^{2}-2-1|\\[0.3em]&=5\cdot |x^{2}-3|\end{aligned}}$

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 $||a|-|b||\leq |a-b|$ :

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots \\[0.3em]&=5\cdot |x^{2}-2|-1|\\[0.3em]&=5\cdot |x^{2}-2|-|-1||\\[0.3em]&\quad {\color {Gray}\left\downarrow \ |a-b|\geq ||a|-|b||\right.}\\[0.3em]&\leq 5\cdot |x^{2}-2-(-1)|\\[0.3em]&=5\cdot |x^{2}-1|\end{aligned}}$

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

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots \\[0.5em]&\leq 5\cdot |x^{2}-1|\\[0.5em]&=5\cdot |x+1||x-1|\end{aligned}}$

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

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots \\[0.5em]&\leq 5\cdot |x^{2}-1|\\[0.5em]&=5\cdot |x+1||x-1|\\[0.5em]&<5\cdot |x+1|\cdot \delta \end{aligned}}$

Now, the required $\delta$ must only depend on $\epsilon$ 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, $\delta \leq 1$ turns out to be quite useful. In fact, $\delta \leq 2$ 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 $\delta \leq 1$ , we now have $|x-1|<1$ and as $x\in [1-1;1+1]$ , we obtain $1+x\in [1;3]$ and $|x+1|\leq 3$. This is the upper bound we were looking for:

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots \\[0.5em]&<5\cdot |x+1|\cdot \delta \\[0.5em]&\quad {\color {Gray}\left\downarrow \ |x+1|\leq 3\right.}\\[0.5em]&\leq 15\cdot \delta \end{aligned}}$

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

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots \\[0.5em]&<15\cdot \delta \\[0.5em]&\leq 15\cdot {\frac {\epsilon }{15}}\\[0.5em]&\leq \epsilon .\end{aligned}}$

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

Proof (Example for a proof of continuity)

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

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

As $\delta =\min \left\{1,{\tfrac {\epsilon }{15}}\right\}$ , there is $\delta \leq 1$. Hence $|x-1|\leq 1$ and $x\in [1-1;1+1]$. It follows that $1+x\in [1;3]$ and therefore $|x+1|\leq 3$.

Step 2: $|f(x)-f(x_{0})|<\epsilon$

${\begin{aligned}|f(x)-f(x_{0})|&=|f(x)-f(1)|\\[0.5em]&=|5\cdot |x^{2}-2|+3-(5|1^{2}-2|+3)|\\[0.5em]&=|5\cdot |x^{2}-2|-5|\\[0.5em]&=5\cdot ||x^{2}-2|-1|\\[0.5em]&\quad {\color {Gray}\left\downarrow \ 1\iff |-1|\right.}\\[0.5em]&=5\cdot ||x^{2}-2|-|-1||\\[0.5em]&\quad {\color {Gray}\left\downarrow \ |a-b|\geq ||a|-|b||\right.}\\[0.5em]&\leq 5\cdot |x^{2}-2-(-1)|\\[0.5em]&=5\cdot |x^{2}-1|\\[0.5em]&\quad {\color {Gray}\left\downarrow \ a^{2}-b^{2}=(a+b)(a-b)\right.}\\[0.5em]&=5\cdot (|x+1|\cdot \underbrace {|x-1|} _{<\delta })\\[0.5em]&<5\cdot |x+1|\cdot \delta \\[0.5em]&\quad {\color {Gray}\left\downarrow \ |x+1|\leq 3\right.}\\[0.5em]&\leq 15\delta \\[0.5em]&\quad {\color {Gray}\left\downarrow \ \delta ={\frac {\epsilon }{15}}\right.}\\[0.5em]&=15\cdot {\frac {\epsilon }{15}}\\[0.5em]&\leq \epsilon \end{aligned}}$

Hence, the function is continuous at $x_{0}=1$ .

Hyperbola

Exercise (Continuity of the hyperbolic function)

Prove that the function $f:\mathbb {R} ^{+}\to \mathbb {R}$ with $f(x)={\tfrac {1}{x}}$ 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 $|x-x_{0}|<\delta \implies |f(x)-f(x_{0})|<\epsilon$ . Forst, let us plug in what we know and reshape our terms a bit until a $|x-x_{0}|$ appears:

${\begin{aligned}|f(x)-f(x_{0})|&=\left|{\frac {1}{x}}-{\frac {1}{x_{0}}}\right|\\[0.5em]&=\left|{\frac {x-x_{0}}{x\cdot x_{0}}}\right|\\[0.5em]&={\frac {|x-x_{0}|}{|x||x_{0}|}}\\[0.5em]&={\frac {1}{|x||x_{0}|}}\cdot |x-x_{0}|\end{aligned}}$

By assumption, there will be $|x-x_{0}|<\delta$, of which we can make use:

$|f(x)-f(x_{0})|=\ldots ={\frac {1}{|x||x_{0}|}}\cdot |x-x_{0}|<{\frac {1}{|x||x_{0}|}}\cdot \delta$

The choice of $\delta$ may again only depend on $\epsilon$ and $x_{0}$ so we need a smart estimate for ${\tfrac {1}{|x|}}$ in order to get rid of the $x$-dependence. To do so, we consider $\delta \leq {\tfrac {x_{0}}{2}}$.

Why was $\delta \leq {\tfrac {x_{0}}{2}}$ and not $\delta \leq 1$ chosen? The explanation is quite simple: We need a $\delta$-neighborhood inside the domain of definition of $f$. If we had simply chosen $\delta \leq 1$ , we might have get kicked out of this domain. For instance, in case$x_{0}={\tfrac {1}{2}}$, the following problem appears:

The biggest $x$-value with $\left|x-{\tfrac {1}{2}}\right|<1$ is $x_{\mathrm {max} }={\tfrac {3}{2}}$ and the smallest one is $x_{\mathrm {min} }=-{\tfrac {1}{2}}$. However, $x_{\mathrm {min} }$ is not inside the domain of definition as $f:\mathbb {R} ^{+}\to \mathbb {R}$. In particular, $x=0$ is element of that interval, where $f(x)={\tfrac {1}{x}}$ 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. $\delta \leq {\tfrac {x_{0}}{2}}$. A third of this distance or other fractions smaller than 1 would also be possible: $\delta \leq {\tfrac {x_{0}}{3}}$, $\delta <{\tfrac {x_{0}}{20}}$ or $\delta <{\tfrac {x_{0}}{5321}}$.

As we chose $\delta \leq {\tfrac {x_{0}}{2}}$ and by $|x-x_{0}|<\delta$ , there is $x\in \left[{\tfrac {x_{0}}{2}};{\tfrac {3x_{0}}{2}}\right]$. This allows for an upper bound: ${\tfrac {1}{|x|}}\leq {\tfrac {2}{|x_{0}|}}$ and we may write:

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots <{\frac {1}{|x||x_{0}|}}\cdot \delta \\[0.5em]&\quad {\color {Gray}\left\downarrow \ \delta ={\frac {\epsilon \cdot |x_{0}|^{2}}{2}}\right.}\\[0.5em]&\leq {\frac {2}{|x_{0}|^{2}}}\cdot \delta \end{aligned}}$

So we get the estimate:

$|f(x)-f(x_{0})|\leq {\frac {2}{|x_{0}|^{2}}}\cdot \delta$

Now, we want to prove $|f(x)-f(x_{0})|<\epsilon$ . Hence we choose $\delta \leq {\tfrac {\epsilon \cdot |x_{0}|^{2}}{2}}$. Plugging this in, our final inequality $|f(x)-f(x_{0})|<\epsilon$ will be fulfilled.

So again, we imposed two conditions for $\delta$ : $\delta \leq {\tfrac {x_{0}}{2}}$ and $\delta \leq {\tfrac {\epsilon \cdot |x_{0}|^{2}}{2}}$. Both are fulfilled by $\delta =\min \left\{{\tfrac {x_{0}}{2}},{\tfrac {\epsilon \cdot |x_{0}|^{2}}{2}}\right\}$, which we will use in the proof:

Proof (Continuity of the hyperbolic function)

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

${\begin{aligned}|f(x)-f(x_{0})|&=\left|{\frac {1}{x}}-{\frac {1}{x_{0}}}\right|\\[0.5em]&=\left|{\frac {x-x_{0}}{x\cdot x_{0}}}\right|\\[0.5em]&={\frac {|x-x_{0}|}{|x||x_{0}|}}\\[0.5em]&={\frac {1}{|x||x_{0}|}}|x-x_{0}|\\[0.5em]&\quad {\color {Gray}\left\downarrow \ {\frac {1}{|x|}}\leq {\frac {2}{|x_{0}|}}\right.}\\[0.5em]&\leq {\frac {2}{|x_{0}|^{2}}}|x-x_{0}|\\[0.5em]&\quad {\color {Gray}\left\downarrow \ |x-x_{0}|<\delta \right.}\\[0.5em]&<{\frac {2\delta }{|x_{0}|^{2}}}\\[0.5em]&\quad {\color {Gray}\left\downarrow \ \delta ={\frac {\epsilon \cdot |x_{0}|^{2}}{2}}\right.}\\[0.5em]&={\frac {2}{|x_{0}|^{2}}}\cdot {\frac {\epsilon \cdot |x_{0}|^{2}}{2}}\\[0.5em]&=\epsilon \end{aligned}}$

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:

$f:\mathbb {R} \to \mathbb {R} ,x\mapsto {\sqrt {5+x^{2}}}$

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 $\epsilon >0$ , there is a $\delta >0$ , such that all $x\in \mathbb {R}$ with $|x-a|<\delta$ satisfy the inequality $|f(x)-f(a)|<\epsilon$ . So let us take a look at the target inequality $|f(x)-f(a)|<\epsilon$ 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

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

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:

${\begin{aligned}|f(x)-f(a)|&=\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\text{expand with}}\left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|\right.}\\[0.3em]&={\frac {\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\cdot \left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}{\left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\geq 0\right.}\\[0.3em]&={\frac {\left|x^{2}-a^{2}\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\cdot |x-a|\\[0.3em]&\quad {\color {Gray}\left\downarrow \ K(x,a):={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\right.}\\[0.3em]&=K(x,a)\cdot |x-a|\end{aligned}}$

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 $K(x,a)\cdot \delta \leq \epsilon$ , then we obtain our target inequality $|f(x)-f(a)|\leq \epsilon$. 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 $K(x,a)\leq {\tilde {K}}(a)$ :

${\begin{aligned}K(x,a)&={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\text{triangle inequality}}\right.}\\[0.3em]&\leq {\frac {\left|x\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}+{\frac {\left|a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\frac {a}{b+c}}\leq {\frac {a}{b}}{\text{ for }}a,c\geq 0,b>0\right.}\\[0.3em]&\leq {\frac {\left|x\right|}{\sqrt {5+x^{2}}}}+{\frac {\left|a\right|}{\sqrt {5+a^{2}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\frac {|a|}{\sqrt {5+a^{2}}}}\leq 1{\text{, since }}{\sqrt {5+a^{2}}}\geq {\sqrt {a^{2}}}=|a|\right.}\\[0.3em]&\leq 2=:{\tilde {K}}(a)\end{aligned}}$

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

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

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

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

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

${\begin{aligned}|f(x)-f(a)|&=\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\\[0.3em]&={\frac {\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\cdot \left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}{\left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}}\\[0.3em]&={\frac {\left|x^{2}-a^{2}\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\cdot |x-a|\\[0.3em]&\leq \left({\frac {\left|x\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}+{\frac {\left|a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\right)\cdot |x-a|\\[0.3em]&\leq \left({\frac {\left|x\right|}{\sqrt {5+x^{2}}}}+{\frac {\left|a\right|}{\sqrt {5+a^{2}}}}\right)\cdot |x-a|\\[0.3em]&\leq (1+1)\cdot |x-a|\\[0.3em]&\leq 2\cdot |x-a|\\[0.3em]&\quad {\color {Gray}\left\downarrow \ |x-a|<\delta ={\frac {\epsilon }{2}}\right.}\\[0.3em]&<2\cdot {\frac {\epsilon }{2}}=\epsilon \end{aligned}}$

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:

$f:\mathbb {R} \to \mathbb {R} :x\mapsto {\begin{cases}\sin \left({\frac {1}{x}}\right)&x\neq 0\\0&x=0\end{cases}}$

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 $\epsilon >0$ and an $x\in \mathbb {R}$, such that $|x-x_{0}|<\delta$ and $|f(x)-f(x_{0})|\geq \epsilon$ . Here, $x$ may be chosen depending on $\delta$ , while $\epsilon$ 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 $|f(x)|\geq \epsilon$.

Step 2: Choose a suitable $\epsilon >0$

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 $\epsilon >0$ , 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 $\epsilon$ from $f(x_{0})=f(0)=0$ . Visually, no matter how small the width $\delta$ of the $2\epsilon$-$2\delta$-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, $\epsilon \leq 1$ may be useful. In that case, there will be function values with $|f(x)|\geq 1$ in every arbitrarily small neighborhood around $x_{0}=0$. We choose $\epsilon ={\tfrac {1}{2}}$. 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 $\epsilon$ 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 $x\in \mathbb {R}$

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

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

So we found several $x$ , where $|f(x)|\geq \epsilon$ . 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 $x={\tfrac {1}{{\tfrac {\pi }{2}}+2k\pi }}$ into this inequality and solve it for $k$ :

${\begin{aligned}\left|x\right|<\delta &\iff \left|{\frac {1}{{\frac {\pi }{2}}+2k\pi }}\right|<\delta \\[0.5em]&\iff {\frac {1}{{\frac {\pi }{2}}+2k\pi }}<\delta \\[0.5em]&\iff 2k\pi +{\frac {\pi }{2}}>{\frac {1}{\delta }}\\[0.5em]&\iff 2k\pi >{\frac {1}{\delta }}-{\frac {\pi }{2}}\\[0.5em]&\iff k>{\frac {1}{2\pi }}\left({\frac {1}{\delta }}-{\frac {\pi }{2}}\right)\end{aligned}}$

So the condition on $k$ is $k>{\tfrac {1}{2\pi }}\left({\tfrac {1}{\delta }}-{\tfrac {\pi }{2}}\right)$. If we choose just any natural number $k$ above this threshold ${\tfrac {1}{2\pi }}\left({\tfrac {1}{\delta }}-{\tfrac {\pi }{2}}\right)$ , 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 $x={\tfrac {1}{{\tfrac {\pi }{2}}+2k\pi }}$ . This gives us both $|x|<\delta$ and $|f(x)|\geq \epsilon$ . So we got all building bricks together, which we will now assemble to a final proof:

Proof (Discontinuity of the topological sine fucntion)

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

${\begin{aligned}k>{\frac {1}{2\pi }}\left({\frac {1}{\delta }}-{\frac {\pi }{2}}\right)&\implies 2k\pi >{\frac {1}{\delta }}-{\frac {\pi }{2}}\\[0.5em]&\implies 2k\pi +{\frac {\pi }{2}}>{\frac {1}{\delta }}\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ {\frac {1}{a}}>{\frac {1}{b}}>0\iff 0<a<b\right.}\\[0.5em]&\implies {\frac {1}{{\frac {\pi }{2}}+2k\pi }}<\delta \\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ {\frac {1}{{\frac {\pi }{2}}+2k\pi }}>0\right.}\\[0.5em]&\implies \left|{\frac {1}{{\frac {\pi }{2}}+2k\pi }}\right|<\delta \\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ x={\frac {1}{{\frac {\pi }{2}}+2k\pi }}\right.}\\[0.5em]&\implies \left|x\right|<\delta \end{aligned}}$

In addition:

${\begin{aligned}\left|f(x)-f(0)\right|&=\left|\sin \left({\frac {1}{x}}\right)-0\right|\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ x={\frac {1}{{\frac {\pi }{2}}+2k\pi }}\right.}\\[0.5em]&=\left|\sin \left({\frac {\pi }{2}}+2k\pi \right)\right|\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ \sin \left({\frac {\pi }{2}}+2k\pi \right)=1\right.}\\[0.5em]&=\left|1\right|\geq {\frac {1}{2}}=\epsilon \end{aligned}}$

Hence, the function is discontinuous at $x_{0}=0$.

Epsilon-Delta Criterion: Hyperbola

Exercise (Continuity of the hyperbolic function)

Prove that the function $f:\mathbb {R} ^{+}\to \mathbb {R}$ with $f(x)={\tfrac {1}{x}}$ 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 $|x-x_{0}|<\delta \implies |f(x)-f(x_{0})|<\epsilon$ . Forst, let us plug in what we know and reshape our terms a bit until a $|x-x_{0}|$ appears:

${\begin{aligned}|f(x)-f(x_{0})|&=\left|{\frac {1}{x}}-{\frac {1}{x_{0}}}\right|\\[0.5em]&=\left|{\frac {x-x_{0}}{x\cdot x_{0}}}\right|\\[0.5em]&={\frac {|x-x_{0}|}{|x||x_{0}|}}\\[0.5em]&={\frac {1}{|x||x_{0}|}}\cdot |x-x_{0}|\end{aligned}}$

By assumption, there will be $|x-x_{0}|<\delta$, of which we can make use:

$|f(x)-f(x_{0})|=\ldots ={\frac {1}{|x||x_{0}|}}\cdot |x-x_{0}|<{\frac {1}{|x||x_{0}|}}\cdot \delta$

The choice of $\delta$ may again only depend on $\epsilon$ and $x_{0}$ so we need a smart estimate for ${\tfrac {1}{|x|}}$ in order to get rid of the $x$-dependence. To do so, we consider $\delta \leq {\tfrac {x_{0}}{2}}$.

Why was $\delta \leq {\tfrac {x_{0}}{2}}$ and not $\delta \leq 1$ chosen? The explanation is quite simple: We need a $\delta$-neighborhood inside the domain of definition of $f$. If we had simply chosen $\delta \leq 1$ , we might have get kicked out of this domain. For instance, in case$x_{0}={\tfrac {1}{2}}$, the following problem appears:

The biggest $x$-value with $\left|x-{\tfrac {1}{2}}\right|<1$ is $x_{\mathrm {max} }={\tfrac {3}{2}}$ and the smallest one is $x_{\mathrm {min} }=-{\tfrac {1}{2}}$. However, $x_{\mathrm {min} }$ is not inside the domain of definition as $f:\mathbb {R} ^{+}\to \mathbb {R}$. In particular, $x=0$ is element of that interval, where $f(x)={\tfrac {1}{x}}$ 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. $\delta \leq {\tfrac {x_{0}}{2}}$. A third of this distance or other fractions smaller than 1 would also be possible: $\delta \leq {\tfrac {x_{0}}{3}}$, $\delta <{\tfrac {x_{0}}{20}}$ or $\delta <{\tfrac {x_{0}}{5321}}$.

As we chose $\delta \leq {\tfrac {x_{0}}{2}}$ and by $|x-x_{0}|<\delta$ , there is $x\in \left[{\tfrac {x_{0}}{2}};{\tfrac {3x_{0}}{2}}\right]$. This allows for an upper bound: ${\tfrac {1}{|x|}}\leq {\tfrac {2}{|x_{0}|}}$ and we may write:

${\begin{aligned}|f(x)-f(x_{0})|&=\ldots <{\frac {1}{|x||x_{0}|}}\cdot \delta \\[0.5em]&\quad {\color {Gray}\left\downarrow \ \delta ={\frac {\epsilon \cdot |x_{0}|^{2}}{2}}\right.}\\[0.5em]&\leq {\frac {2}{|x_{0}|^{2}}}\cdot \delta \end{aligned}}$

So we get the estimate:

$|f(x)-f(x_{0})|\leq {\frac {2}{|x_{0}|^{2}}}\cdot \delta$

Now, we want to prove $|f(x)-f(x_{0})|<\epsilon$ . Hence we choose $\delta \leq {\tfrac {\epsilon \cdot |x_{0}|^{2}}{2}}$. Plugging this in, our final inequality $|f(x)-f(x_{0})|<\epsilon$ will be fulfilled.

So again, we imposed two conditions for $\delta$ : $\delta \leq {\tfrac {x_{0}}{2}}$ and $\delta \leq {\tfrac {\epsilon \cdot |x_{0}|^{2}}{2}}$. Both are fulfilled by $\delta =\min \left\{{\tfrac {x_{0}}{2}},{\tfrac {\epsilon \cdot |x_{0}|^{2}}{2}}\right\}$, which we will use in the proof:

Proof (Continuity of the hyperbolic function)

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

${\begin{aligned}|f(x)-f(x_{0})|&=\left|{\frac {1}{x}}-{\frac {1}{x_{0}}}\right|\\[0.5em]&=\left|{\frac {x-x_{0}}{x\cdot x_{0}}}\right|\\[0.5em]&={\frac {|x-x_{0}|}{|x||x_{0}|}}\\[0.5em]&={\frac {1}{|x||x_{0}|}}|x-x_{0}|\\[0.5em]&\quad {\color {Gray}\left\downarrow \ {\frac {1}{|x|}}\leq {\frac {2}{|x_{0}|}}\right.}\\[0.5em]&\leq {\frac {2}{|x_{0}|^{2}}}|x-x_{0}|\\[0.5em]&\quad {\color {Gray}\left\downarrow \ |x-x_{0}|<\delta \right.}\\[0.5em]&<{\frac {2\delta }{|x_{0}|^{2}}}\\[0.5em]&\quad {\color {Gray}\left\downarrow \ \delta ={\frac {\epsilon \cdot |x_{0}|^{2}}{2}}\right.}\\[0.5em]&={\frac {2}{|x_{0}|^{2}}}\cdot {\frac {\epsilon \cdot |x_{0}|^{2}}{2}}\\[0.5em]&=\epsilon \end{aligned}}$

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:

$f:\mathbb {R} \to \mathbb {R} ,x\mapsto {\sqrt {5+x^{2}}}$

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 $\epsilon >0$ , there is a $\delta >0$ , such that all $x\in \mathbb {R}$ with $|x-a|<\delta$ satisfy the inequality $|f(x)-f(a)|<\epsilon$ . So let us take a look at the target inequality $|f(x)-f(a)|<\epsilon$ 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

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

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:

${\begin{aligned}|f(x)-f(a)|&=\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\text{expand with}}\left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|\right.}\\[0.3em]&={\frac {\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\cdot \left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}{\left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\geq 0\right.}\\[0.3em]&={\frac {\left|x^{2}-a^{2}\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\cdot |x-a|\\[0.3em]&\quad {\color {Gray}\left\downarrow \ K(x,a):={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\right.}\\[0.3em]&=K(x,a)\cdot |x-a|\end{aligned}}$

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 $K(x,a)\cdot \delta \leq \epsilon$ , then we obtain our target inequality $|f(x)-f(a)|\leq \epsilon$. 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 $K(x,a)\leq {\tilde {K}}(a)$ :

${\begin{aligned}K(x,a)&={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\text{triangle inequality}}\right.}\\[0.3em]&\leq {\frac {\left|x\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}+{\frac {\left|a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\frac {a}{b+c}}\leq {\frac {a}{b}}{\text{ for }}a,c\geq 0,b>0\right.}\\[0.3em]&\leq {\frac {\left|x\right|}{\sqrt {5+x^{2}}}}+{\frac {\left|a\right|}{\sqrt {5+a^{2}}}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\frac {|a|}{\sqrt {5+a^{2}}}}\leq 1{\text{, since }}{\sqrt {5+a^{2}}}\geq {\sqrt {a^{2}}}=|a|\right.}\\[0.3em]&\leq 2=:{\tilde {K}}(a)\end{aligned}}$

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

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

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

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

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

${\begin{aligned}|f(x)-f(a)|&=\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\\[0.3em]&={\frac {\left|{\sqrt {5+x^{2}}}-{\sqrt {5+a^{2}}}\right|\cdot \left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}{\left|{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}\right|}}\\[0.3em]&={\frac {\left|x^{2}-a^{2}\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\\[0.3em]&={\frac {\left|x+a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\cdot |x-a|\\[0.3em]&\leq \left({\frac {\left|x\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}+{\frac {\left|a\right|}{{\sqrt {5+x^{2}}}+{\sqrt {5+a^{2}}}}}\right)\cdot |x-a|\\[0.3em]&\leq \left({\frac {\left|x\right|}{\sqrt {5+x^{2}}}}+{\frac {\left|a\right|}{\sqrt {5+a^{2}}}}\right)\cdot |x-a|\\[0.3em]&\leq (1+1)\cdot |x-a|\\[0.3em]&\leq 2\cdot |x-a|\\[0.3em]&\quad {\color {Gray}\left\downarrow \ |x-a|<\delta ={\frac {\epsilon }{2}}\right.}\\[0.3em]&<2\cdot {\frac {\epsilon }{2}}=\epsilon \end{aligned}}$

Hence, $f$ is a continuous function.