# Math for Non-Geeks/ Sequential definition of continuity

Among the sequence criteria, the epsilon-delta criterion is another way to define the continuity of functions. This criterion describes the feature of continuous functions, that sufficiently small changes of the argument cause arbitrarily small changes of the function value.

## Motivation

In the beginning of this chapter, we learned that continuity of a function may - by a simple intuition - be considered as an absence of jumps. So if we are at an argument where continuity holds, the function values will change arbitrarily little, when we wiggle around the argument by a sufficiently small amount. So $f(x)\approx f(x_{0})$, for $x$ in the vicinity of $x_{0}$ . The function values $f(x)$ may therefore be useful to approximate $f(x_{0})$ .

### Continuity when approximating function values

If a function has no jumps, we may approximate its function values by other nearby values . For this approximation, and hence also for proofs of continuity, we will use the epsilon-delta criterion for continuity. So how will such an approximation look in a practical situation?
Suppose, we make an experiment that includes measuring the air temperature as a function of time. Let $f$ be the function describing the temperature. So $f(x)$ is the temperature at time $x$. Now, suppose there is a technical problem, so we have no data for $f(x_{0})$ - or we simply did not measure $f$ at exactly this point of time. However, we would like to approximate the function value $f(x_{0})$ as precisely as we can:
Suppose, a technical issue prevented the measurement of $f(x_{0})$ . Since the temperature changes continuously in time - and especially there is no jump at $x_{0}$ - we may instead use a temperature value measured at a time close to $x_{0}$ . So, let us approximate the value $f(x_{0})$ by taking a temperature $f(x)$ with $x$ close to $x_{0}$ . That means, $f(x)$ is an approximation for $f(x_{0})$. How close must $x$ come to $x_{0}$ in order to obtain a given approximation precision?
Suppose that for the evaluation of the temperature at a later time $x$ , the maximal error shall be $\epsilon =0{.}1\ \mathrm {^{\circ }C}$ . So considering the following figure, the measured temperature should be in the grey region . Those are all temperatures with function values between $f(x_{0})-\epsilon$ and $f(x_{0})+\epsilon$ , i.e. inside the open interval $(f(x_{0})-\epsilon ,f(x_{0})+\epsilon )$ :
In this graphic, we may see that there is a region around $x_{0}$ , where function values differ by less than $\epsilon$ from $f(x_{0})$ . So in fact, there is a time difference $\delta$, such that all function values are inside the interval $(x_{0}-\delta ,x_{0}+\delta )$ highlighted in grey:
Therefore, we may indeed approximate the missing data point $f(x_{0})$ sufficiently well (meaning with a maximal error of $\epsilon$) . This is done by taking a time $x$ differing from $x_{0}$ by less than $\delta$ and then, the error of $f(x)$ in approximating $f(x_{0})$ will be smaller than the desired maximal error $\epsilon$. So $f(x)$ will be the approximation for $f(x_{0})$ .

### Increasing approximation precision

What will happen, if we need to know the temperature value to a higher precision due to increased requirements in the evaluation of the experiment? For instance, if the required maximal temperature error is set to $\epsilon _{2}=0{.}05\ \mathrm {^{\circ }C}$ instead of $\epsilon =0{.}1\ \mathrm {^{\circ }C}$ ?
In that case, thare is an interval around $x_{0}$, where function values do not deviate by more than $\epsilon _{2}$ from $f(x_{0})$ . Mathematically speaking, there a $\delta _{2}>0$ exists, such that $f(x)$ differs by a maximum amount of $\epsilon _{2}$ from $f(x_{0})$ , if there is $|x-x_{0}|<\delta _{2}$ :
No matter how small we choose $\epsilon$ , thanks to the continuous temperature dependence, we may always find a $\delta >0$ , such that $f(x)$ differs at most by $\epsilon$ from $f(x_{0})$ , whenever $x$ is closer to $x_{0}$ than $\delta$ . We keep in mind:
This holds true , since the function $f$ does not have a jump at $x_{0}$ . In other words, since $f$ is continuous at $x_{0}$. Even beyond that, we may always infer from the above characteristic that there is no jump in the graph of $f$ at $x_{0}$. Therefore, we may use it as a formal definition for continuity. As mathematicians frequently use the variables $\epsilon$ and $\delta$ when describing this characteristic, it is also called epsilon-delta-criterion for continuity.

### Epsilon-delta-criterion for continuity

Why does the epsilon-delta-criterion hold if and only if the graph of the function does not have a jump at some argument (i.e. it is continuous there)? The temperature example allows us to intuitively verify, that the epsilon-delta-criterion is satisfied for continuous functions. But will the epsilon-delta-criterion be violated, when a function has a jump at some argument? To answer this question, let us assume that the temperature as a function of time has a jump at some $x_{0}$:
Let $\epsilon$ be a given maximal error that is smaller than the jump:
In that case, we may not choose a $\delta$-interval $(x_{0}-\delta ,x_{0}+\delta )$ around $x_{0}$, where all function values have a deviation lower than $\epsilon$ from $f(x_{0})$. If we, for instance, choose the following $\delta$, then there certainly is an $x$ between $x_{0}-\delta$ and $x_{0}+\delta$ with a function value differing by more than $\epsilon$ from $f(x_{0})$:
When choosing a smaller $\delta _{2}$, we will find an $x\in (x_{0}-\delta _{2},x_{0}+\delta _{2})$ with $|f(x)-f(x_{0})|\geq \epsilon$, as well:
No matter how small we choose $\delta$, there will always be an argument $x$ with a distance of less than $\delta$ to $x_{0}$, such that the function value $f(x)$ differs by more than $\epsilon$ from $f(x_{0})$. So we have seen that in an intuitive example, the epsilon-delta-criterion is not satisfied, if the function has a jump. Therefore, the epsilon-delta-criterion characterizes whether the graph of the function has a jump at the considered argument $x_{0}$ or not. That means, we may consider it as a definition of continuity. Since this criterion only uses mathematically well-defined terms, it may be used not just as an intuitive, but also as a formal definition.
Hint
In the above example, we did not pay attention to some of the aspects we would have to consider when performing actual measurements. As an example, we assumed to have perfect measurements without any errors. Of course, this is not the case in reality. Every measurement at some time $x$ has an outcome differing from the actual value $f(x)$. In addition, we assumed instantaneous measurements at $x$. Usually, each recording of a value takes some time. These uncertainties have to be taken into account for real experiments.

## Definition
### Epsilon-Delta criterion for continuity

The $\epsilon$-$\delta$ definition of continuity at an argument $x_{0}$ inside the domain of definition is the following:
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
$\forall \epsilon >0\,\exists \delta >0\,\forall x\in D:|x-x_{0}|<\delta \implies |f(x)-f(x_{0})|<\epsilon$
Explanation of the quantifier notation:
${\begin{aligned}{\begin{array}{l}\underbrace {{\underset}\forall \epsilon >0} _{{\text{For all }}\epsilon >0}\underbrace {{\underset}\exists \delta >0} _{{\text{ there is a }}\delta >0}\underbrace {{\underset}\forall x\in D} _{{\text{, such that for all }}x\in D}\\[1em]\quad \underbrace {{\underset}|x-x_{0}|<\delta } _{{\text{ with distance to }}x_{0}{\text{ smaller than }}\delta }\underbrace {{\underset}\implies } _{\text{ there is}}\underbrace {{\underset}|f(x)-f(x_{0})|<\epsilon } _{{\text{, such that the distance of }}f(x){\text{ to }}f(x_{0}){\text{ is smaller than }}\epsilon }\end{array}}\end{aligned}}$
The above definition describes continuity at a certain point (argument). An entire function $f:D\to \mathbb {R}$ is called continuous, when it is continuous - according to the epsilon-delta criterion - at each of its arguments in the domain of definition.

### Derivation of the Epsilon-Delta criterion for discontinuity

We may also obtain a criterion of discontinuity by simply negating the above definition. Negating mathematical propositions has already been treated in chapter „Aussagen negieren“ . While doing so, an all quantifier $\forall$ gets transformed into an existential quantifier $\exists$ and vice versa. Concerning inner implication, we have to keep in mind that the negation of $A\implies B$ is equivalent to $A\land \neg B$ . Negating the epsilon-delta criterion of discontinuity, we obtain:
${\begin{aligned}{\begin{array}{rrrrrcr}&\neg {\Big\forall \epsilon >0\,&\exists \delta >0\,&\forall x\in D:&|x-x_{0}|<\delta &\implies &|f(x)-f(x_{0})|<\epsilon {\Big )}\\[0.5em]\iff &\exists \epsilon >0\,&\neg {\Big\exists \delta >0\,&\forall x\in D:&|x-x_{0}|<\delta &\implies &|f(x)-f(x_{0})|<\epsilon {\Big )}\\[0.5em]\iff &\exists \epsilon >0\,&\forall \delta >0\,&\neg {\Big\forall x\in D:&|x-x_{0}|<\delta &\implies &|f(x)-f(x_{0})|<\epsilon {\Big )}\\[0.5em]\iff &\exists \epsilon >0\,&\forall \delta >0\,&\exists x\in D:&\neg {\Big|x-x_{0}|<\delta &\implies &|f(x)-f(x_{0})|<\epsilon {\Big )}\\[0.5em]\iff &\exists \epsilon >0\,&\forall \delta >0\,&\exists x\in D:&|x-x_{0}|<\delta &\land &\neg {\Big|f(x)-f(x_{0})|<\epsilon {\Big )}\\[0.5em]\iff &\exists \epsilon >0\,&\forall \delta >0\,&\exists x\in D:&|x-x_{0}|<\delta &\land &|f(x)-f(x_{0})|\geq \epsilon \end{array}}\end{aligned}}$
This gets us the negation of continuity (i.e. discontinuity):
$\exists \epsilon >0\,\forall \delta >0\,\exists x\in D:|x-x_{0}|<\delta \land |f(x)-f(x_{0})|\geq \epsilon$

### Epsilon-Delta criterion for discontinuity

Definition (Epsilon-Delta definition of discontinuity)
A function $f:D\to \mathbb {R}$ with $D\subseteq \mathbb {R}$ is discontinuous at $x_{0}\in D$, if and only if there is an $\epsilon >0$ , such that for all $\delta >0$ a $x\in D$ with $|x-x_{0}|<\delta$ and $|f(x)-f(x_{0})|\geq \epsilon$ exists. Mathematically written, $f$ is discontinuous at $x_{0}\in D$ iff
$\exists \epsilon >0\,\forall \delta >0\,\exists x\in D:|x-x_{0}|<\delta \land |f(x)-f(x_{0})|\geq \epsilon$
Explanation of the quantifier notation:
${\begin{aligned}{\begin{array}{l}\underbrace {{\underset}\exists \epsilon >0} _{{\text{There is a }}\epsilon >0,}\underbrace {{\underset}\forall \delta >0} _{{\text{ such that for all }}\delta >0}\underbrace {{\underset}\exists x\in D} _{{\text{ there is an }}x\in D}\\[1em]\quad \underbrace {{\underset}|x-x_{0}|<\delta } _{{\text{ with distance to }}x_{0}{\text{ smaller than }}\delta }\underbrace {{\underset}\land } _{\text{ and}}\underbrace {{\underset}|f(x)-f(x_{0})|\geq \epsilon } _{{\text{the distance of }}f(x){\text{ to }}f(x_{0}){\text{ is bigger (or equal) }}\epsilon }\end{array}}\end{aligned}}$

## Further explanations considering the Epsilon-Delta criterion

The inequality $|x-x_{0}|<\delta$ means that the distance between $x$ and $x_{0}$ is smaller than $\delta$ . Analogously, $|f(x)-f(x_{0})|<\epsilon$ tells us that the distance between $f(x)$ and $f(x_{0})$ is smaller than $\epsilon$ . Therefor, the implication $|x-x_{0}|<\delta \implies |f(x)-f(x_{0})|<\epsilon$ just says that whenever $f(x)$ and $f(x_{0})$ are closer together than $\epsilon$ , then we know that the distance between $x$ and $x_{0}$ before applying the function must have been smaller than $\delta$ . Thus we may interpret the epsilon-delta criterion in the following way:
For continuous functions, we can control the error $f(x)$ to be lower than $\epsilon$ by keeping the error in the argument sufficiently small (smaller than $\delta$). Finding a $\delta$ means answering the question: How low does my initial error in the argument have to be in order to get a final error smaller than $\epsilon$ . This may get interesting when doing numerical calculations or measurements. Imagine, you are measuring some $x_{0}$ and then using it to compute $f(x_{0})$ where $f$ is a continuous function. The epsilon-delta criterion allows you to find the maximal error $\delta$ in $x$ (i.e. $|x-x_{0}|<\delta$), which guarantees that the final error $|f(x)-f(x_{0})|$ will be smaller than $\epsilon$.
A $\delta$ may only be found if small changes around the argument $x_{0}$ also cause small changes around the function value $f(x_{0})$ . Hence, concerning functions continuous at $x_{0}$ , there has to be:
$x\approx x_{0}\implies f(x)\approx f(x_{0})$
I.e.: whenever $x$ is sufficiently close to $x_{0}$ , then $f(x)$ is approximately $f(x_{0})$. This may also be described using the notion of an $\epsilon$-neighborhood:
In topology, this description using neighborhoods will be generalized to a topological definition of continuity.

## Visualization of the Epsilon-Delta criterion
### Description of continuity using the graph

The epsilon-delta criterion may nicely be visualized by taking a look at the graph of a funtion. Let's start by getting a picture of the implication $|x-x_{0}|<\delta \implies |f(x)-f(x_{0})|<\epsilon$. This means, the distance between $f(x)$ and $f(x_{0})$ is smaller than epsilon, whenever $x$ is closer to $x_{0}$ than $\delta$ . So for $x\in (x_{0}-\delta ,x_{0}+\delta )$, there is $f(x)\in (f(x_{0})-\epsilon ,f(x_{0})+\epsilon )$. Hence, the point $(x,f(x))$ has to be inside the rectangle $(x_{0}-\delta ,x_{0}+\delta )\times (f(x_{0})-\epsilon ,f(x_{0})+\epsilon )$ . This is a rectangle with width $2\delta$ and height $2\epsilon$ centered at $(x_{0},f(x_{0}))$:
We will call this the $2\epsilon$-$2\delta$-rectangle and only consider its interior. That means, the boundary does not belong to the rectangle. Following the epsilon-delta criterion, the implication $|x-x_{0}|<\delta \implies |f(x)-f(x_{0})|<\epsilon$ has to be fulfilled for all arguments $x$ . Thus, all points making up the graph of $f$ restricted to arguments inside the interval $(x_{0}-\delta ,x_{0}+\delta )$ (in the interior of the $2\epsilon$-$2\delta$-rectangle, which is marked green) must never be above or below the rectangle (the red area):
So graphically, we may describe the epsilon-delta criterion as follows:

### Example of a continuous function

For an example, consider the function $f:\mathbb {R} \to \mathbb {R} :x\mapsto {\tfrac {1}{3}}x$ . This fucntion is continuous everywhere - and hence also at the argument $x_{0}=1$. There is $f(x_{0})=f(1)={\tfrac {1}{3}}\cdot 1={\tfrac {1}{3}}$. At first, consider a maximal final error of $\epsilon =1$ around $f(x_{0})$. With $\delta =2$ , we can find a $\delta >0$, such that the graph of $f$ is entirely situated inside the interior of the $2\epsilon$-$2\delta$-rectangle:
But not only for $\epsilon =1$, but for any $\epsilon >0$ we may find a $\delta >0$ , such that the graph of $f$ is situated entirely inside the respective $2\epsilon$-$2\delta$-rectangle:
* For $\epsilon ={\tfrac {1}{2}}$ , one can choose $\delta =1$ and the graph is in the interior of the $2\epsilon$-$2\delta$-rectangle.
* In case $\epsilon ={\tfrac {1}{3}}$, the width $\delta ={\tfrac {3}{4}}$ will be small enough to get the graph into the $2\epsilon$-$2\delta$-rectangle.

### Example for a discontinuous function

What happens if the function is discontinuous? Let's take the signum function $\operatorname {sgn}$, which is discontinuous at 0:
$\operatorname {sgn} :\mathbb {R} \to \mathbb {R} :x\mapsto {\begin{cases}1&x>0\\0&x=0\\-1&x<0\end{cases}}$
And here is its graph:
The graph intuitively allows to recognize that at $x_{0}=0$ , there certainly is a discontinuity. And we may see this using the rectangle visualization, as well. When choosing a rectangle height $\epsilon$, smaller than the jump height (i.e. $\epsilon <1$), then there is no $\delta$, such that the graph can be fitted entirely inside the $2\epsilon$-$2\delta$-rectangle. For instance if $\epsilon ={\tfrac {1}{2}}$ , then for any $\delta$ - no matter how small - there will always be function values above or below the $2\epsilon$-$2\delta$-rectangle. In fact, this apples to all values except for $f(0)=0$:
* For $\epsilon ={\tfrac {1}{2}}$ and $\delta =2$ , the signum function has values above or below the $2\epsilon$-$2\delta$-rectangle (colored in red).
* For $\delta ={\tfrac {1}{2}}$ we will find points in the graph above or below the $2\epsilon$-$2\delta$-rectangle, as well.

## Dependence of delta or epsilon choice
### Continuity

How does the choice of $\delta >0$ depend on $x_{0}$ and $\epsilon$? Suppose, an arbitrary $\epsilon >0$ is given in order to check continuity of $f$. Now, we need to find a rectangle width $\delta >0$ , such that the restriction of the graph of $f$ to arguments inside the interval $(x_{0}-\delta ,x_{0}+\delta )$ entirely fits into the epsilon-tube $(f(x_{0})-\epsilon ,f(x_{0})+\epsilon )$ . This of course requires choosing $\delta$sufficiently small. When $\delta$ is too large, there may be an argument $x$ in $(x_{0}-\delta ,x_{0}+\delta )$, where $f(x)$ has escaped the tube, i.e. it has a distance to $f(x_{0})$ larger than $\epsilon$ :
* If for a given $\epsilon$ , the respective $\delta$ is chosen too large, then there may be function values above or below the $2\epsilon$-$2\delta$-rectangle (marked red, here).
* By contrast, if $\delta$ is re-scaled to be sufficiently small, the graph entirely fits into the $2\epsilon$-$2\delta$-rectangle.
How small $\delta$ has to be chosen, will depend on three factors: The function $f$, the given $\epsilon$ and the argument $x_{0}$. Depending on the function slope, a different $\delta$ chosen (steep functions require a smaller $\delta$). Furthermore, for a smaller $\epsilon$ we also have to choose a smaller $\delta$ . The following diagrams illustrate this: Here, a quadratic function is plotted, which is continuous at $x_{0}=1$ . For a smaller $\epsilon$ , we also need to choose a smaller $\delta$ :
The choice of $\delta$ will depend on the argument $x_{0}$, as well. The more a function changes in the neighborhood of a certain point (i.e. it is steep around it), the smaller we have to choose $\delta$ . The following graphic demonstrates this: The $\delta$-value proposed there is sufficiently small at $x_{0}$ , but too large at $x_{1}$ :
In the vicinity of $x_{1}$ , the function $f$ has a higher slope compared to $x_{0}$. Hence, we need to choose a smaller $\delta$ at $x_{1}$ . Let us denote the $\delta$-values at $x_{0}$ and $x_{1}$ correspondingly by $\delta _{0}$ and $\delta _{1}$ - and choose $\delta _{1}$ to be smaller:
So, we have just seen that the choice of $\delta$ depends on the function $f$ to be considered, as well as the argument $x_{0}$ and the given $\epsilon$ .

### Discontinuity

For a discontinuity proof, the relations between the variables will interchange. This relates back to the interchange of the quantifiers under negation of propositions. In order to show discontinuity, we need to find an $\epsilon >0$ small enough, such that for no $\delta >0$ the graph of $f$ fits entirely into the $2\epsilon$-$2\delta$-rectangle. In particular, if the discontinuity is caused by a jump, then $\epsilon$ must be chosen smaller than the jump height. For $\epsilon$ too large, there might be a $\delta$, such that $f$ does fit into the $2\epsilon$-$2\delta$-rectangle:
* Choosing $\epsilon$ too lage for the signum function, we get a $\delta$, such that the graph entirely fits into the $2\epsilon$-$2\delta$-rectangle.
* If $\epsilon$ is chosen small enough, then for any $\delta >0$ there will be function values above or below the $2\epsilon$-$2\delta$-retangle.
Which $\epsilon$has to be chosen again depends on the function around $x_{0}$ . After $\epsilon$ has been chosen, an arbitrary $\delta >0$ will be considered. Then, an $x$ between $x_{0}-\delta$ and $x_{0}+\delta$ has to be found, such that $f(x)$ has a distance larger than (or equal to) $\epsilon$ to $f(x_{0})$ . That means, the point $(x,f(x))$ has to be situated above or below the $2\epsilon$-$2\delta$-rectangle. Which $x$ has to be chosen depends on a varety of parameters: the chosen $\epsilon$ and the arbitrarily given $\delta$, the discontinuity and the behavior of the function around it.

## Example problems
### Continuity

Math for Non-Geeks: Template:Aufgabe

### Discontinuity

Math for Non-Geeks: Template:Aufgabe

## Relation to the sequence criterion

Now, we have two definitions of continuity: the epsilon-delta and the sequence criterion. In order to show that both definitions describe the same concept, we have to prove their equivalence. If the sequence criterion is fulfilled, it must imply that the epsilon-delta criterion holds and vice versa.

### Epsilon-delta criterion implies sequence criterion

Theorem (The epsilon-delta criterion implies the sequence criterion)
Let $f:D\to \mathbb {R}$ with $D\subseteq \mathbb {R}$ be any function. If this function satisfies the epsilon-dela criterion at $x_{0}\in D$ , then the sequence criterion is fulfilled at $x_{0}$ , as well.
How to get to the proof? (The epsilon-delta criterion implies the sequence criterion)
Let us assume that the function $f:D\to \mathbb {R}$ satisfies the epsilon-delta criterion at $x_{0}\in D$ . That means:
We now want to prove that the sequence criterion is satisfied, as well. So we have to show that for any sequence of arguments $(x_{n})_{n\in \mathbb {N} }$ converging to $x_{0}$ , there also has to be $\lim _{n\to \infty }f(x_{n})=f(x_{0})$ . We therefor consider an arbitrary sequence of arguments $(x_{n})_{n\in \mathbb {N} }$ in the domain $x_{n}\in D$ with $\lim _{n\to \infty }x_{n}=x_{0}$. Our job is to show that the sequence of function values $\left(f(x_{n})\right)_{n\in \mathbb {N} }$ converges to $f(x_{0})$ . So by the definition of convergence:
Let $\epsilon >0$ be arbitrary. We have to find a suitable $N\in \mathbb {N}$ with $|f(x_{n})-f(x_{0})|<\epsilon$ for all sequence elements beyond that $N$ , i.e. $n\geq N$ . The inequality $|f(x_{n})-f(x_{0})|<\epsilon$ seems familiar, recalling the epsilon-delta criterion. The only difference is that the argument $x$ is replaced by a sequence element $x_{n}$ - so we consider a special case for $x$ . Let us apply the epsilon-delta criterion to that special case, with our arbitrarily chosen $\epsilon$ being given:
Our goal is coming closer. Whenever a sequence element $x_{n}$ is close to $x_{0}$ with $|x_{n}-x_{0}|<\delta$ , it will satisfy the inequality which we want to show, namely $|f(x_{n})-f(x_{0})|<\epsilon$. It remains to choose an $N\in \mathbb {N}$, where this is the case for all sequence elements beyond $x_{N}$ . The convergence $\lim _{n\to \infty }x_{n}=x_{0}$ implies that $|x_{n}-x_{0}|$ gets arbitrarily small. So by the definition of continuity, we may find an ${\tilde {N}}\in \mathbb {N}$, with $|x_{n}-x_{0}|<\delta$ for all $n\geq {\tilde {N}}$ . This ${\tilde {N}}$ now plays the role of our $N$. If there is $n\geq N={\tilde {N}}$, it follows that $|x_{n}-x_{0}|<\delta$ and hence $|f(x_{n})-f(x_{0})|<\epsilon$ by the epsilon-delta criterion. In fact, any $N\geq {\tilde {N}}$ will do the job. We now conclude our considerations and write down the proof:
Proof (The epsilon-delta criterion implies the sequence criterion)
Let $f:D\to \mathbb {R}$ e a function satisfying the epsilon-delta criterion at $x_{0}\in D$ . Let $(x_{n})_{n\in \mathbb {N} }$ be a sequence inside the domain of definition, i.e. $x_{n}\in D$ for all $n\in \mathbb {N}$ coverging as $\lim _{n\to \infty }x_{n}=x_{0}$. We would like to show that for any given $\epsilon >0$ there exists an $N\in \mathbb {N}$ , such that $|f(x_{n})-f(x_{0})|<\epsilon$ holds for all $n\geq N$ .
So let $\epsilon >0$ be given. Following the epsilon-delta criterion, there is a $\delta >0$, with $|f(x)-f(x_{0})|<\epsilon$ for all $x\in D$ close to $x_{0}$ , i.e. $|x-x_{0}|<\delta$ . As $(x_{n})_{n\in \mathbb {N} }$ converges to $x_{0}$ , we may find an $N\in \mathbb {N}$ with $|x_{n}-x_{0}|<\delta$ for all $n\geq N$ .
Now, let $n\geq N$ be arbitrary. Hence, $|x_{n}-x_{0}|<\delta$. The epsilon-delta criterion now implies $|f(x_{n})-f(x_{0})|<\epsilon$. This proves $\lim _{n\to \infty }f(x_{n})=f(x_{0})$ and therefore establishes the epsilon-delta criterion.

### Sequence criterion implies epsilon-delta criterion

Theorem (The sequence criterion implies the epsilon-delta criterion)
Let $f:D\to \mathbb {R}$ with $D\subseteq \mathbb {R}$ be a function. If $f$ satisfies the sequence criterion at $x_{0}\in D$ , then the epsilon-delta criterion is fulfilled there, as well.
How to get to the proof? (The sequence criterion implies the epsilon-delta criterion)
We need to show that the following implication holds:
$f{\text{ satisfies the sequence criterion in }}x_{0}\in D\implies f{\text{ satisfies the epsilon-delta criterion in }}x_{0}\in D$
This time, we do not show the implication directly, but using a contraposition. So we will prove the following implication (which is equivalent to the first one):
$\neg \left(f{\text{ satisfies the epsilon-delta criterion in }}x_{0}\in D\right)\implies \neg \left(f{\text{ satisfies the sequence criterion in }}x_{0}\in D\right)$
Or in other words:
$f{\text{ violates the epsilon-delta criterion in }}x_{0}\in D\implies f{\text{ violates the sequence criterion in }}x_{0}\in D$
So let $f:D\to \mathbb {R}$ be a function that violates the epsilon-delta criterion at $x_{0}\in D$ . Hence, $f$ fulfills the discontinuity version of the epsilon-delta criterion at $x_{0}$. We can find an $\epsilon >0$,such that for any $\delta >0$ there is a $x\in D$ with $|x-x_{0}|<\delta$ but $|f(x)-f(x_{0})|\geq \epsilon$ . It is our job now to prove, that the sequence criterion is violated, as well. This requires choosing a sequence of aguments $(x_{n})_{n\in \mathbb {N} }$ , converging as $\lim _{n\to \infty }x_{n}=x_{0}$ but $\lim _{n\to \infty }f(x_{n})\neq f(x_{0})$ .
This choice will be done exploiting the discontinuity version of the epsilon-delta criterion. That version provides us with an $\epsilon >0$ , where $|f(x)-f(x_{0})|\geq \epsilon$ holds (so continuity is violated) for certain arguments $x$ . We will now construct our sequence exclusively out of those certain $x$ . This will automatically get us $\lim _{n\to \infty }f(x_{n})\neq f(x_{0})$.
So how to find a suitable sequence of arguments $(x_{n})_{n\in \mathbb {N} }$, converging to $x_{0}$ ? The answer is: by choosing a null sequence $(\delta _{n})_{n\in \mathbb {N} }$. Practically, this is done as follows: we set $\delta _{n}={\tfrac {1}{n}}$ . For any $\delta _{n}$ , we take one of the certain $x$ for $\delta =\delta _{n}$ as our argument $x_{n}$ . Then, $|x_{n}-x_{0}|<\delta _{n}$ but also $|f(x_{n})-f(x_{0})|\geq \epsilon$. These $x_{n}$ make up the desired sequence $(x_{n})_{n\in \mathbb {N} }$. On one hand, there is $|x_{n}-x_{0}|<\delta _{n}$ and as $\lim _{n\to \infty }\delta _{n}=0$ , the convergence $\lim _{n\to \infty }x_{n}=x_{0}$ holds. But on the other hand $|f(x_{n})-f(x_{0})|\geq \epsilon$ , so the sequence of function values $\left(f(x_{n})\right)_{n\in \mathbb {N} }$ does not converge to $f(x_{0})$ . Let us put these thoughts together in a single proof:
Proof (The sequence criterion implies the epsilon-delta criterion)
We establish the theorem by contraposition. It needs to be shown that a function $f:D\to \mathbb {R}$ violating the epsilon-delta criterion at $x_{0}\in D$ also violates the sequence criterion at $x_{0}$ . So let $f:D\to \mathbb {R}$ with $D\subseteq \mathbb {R}$ be a function violating the epsilon-delta criterion at $x_{0}\in D$ . Hence, there is an $\epsilon >0$, such that for all $\delta >0$ an $x\in D$ exists with $|x-x_{0}|<\delta$ but $|f(x)-f(x_{0})|\geq \epsilon$ .
So for any $\delta _{n}={\tfrac {1}{n}}$ , there is an $x_{n}\in D$ with $|x_{n}-x_{0}|<\delta _{n}$ but $|f(x_{n})-f(x_{0})|\geq \epsilon$. The inequality $|x_{n}-x_{0}|<\delta _{n}$ can also be written $x_{0}-\delta _{n}\leq x_{n}\leq x_{0}+\delta _{n}$. As $\lim _{n\to \infty }\delta _{n}=0$ , there is both $\lim _{n\to \infty }x_{0}-\delta _{n}=x_{0}$ and $\lim _{n\to \infty }x_{0}+\delta _{n}=x_{0}$. Thus, by the sandwich theorem, the sequence $(x_{n})_{n\in \mathbb {N} }$ converges to $x_{0}$.
But since $|f(x_{n})-f(x_{0})|\geq \epsilon$ for all $n\in \mathbb {N}$ , the sequence $\left(f(x_{n})\right)_{n\in \mathbb {N} }$ can not converge to $f(x_{0})$ . Therefore, the sequence criterion is violated at $x_{0}$ for the function $f$ : We have found a sequence of arguments $(x_{n})_{n\in \mathbb {N} }$ with $\lim _{n\to \infty }x_{n}=x_{0}$ but $\lim _{n\to \infty }f(x_{n})\neq f(x_{0})$.

## Exercises
### Quadratic function

Math for Non-Geeks: Template:Aufgabe

### Concatenated absolute function

Math for Non-Geeks: Template:Aufgabe

### Hyperbola

Math for Non-Geeks: Template:Aufgabe

### Concatenated square root function

Math for Non-Geeks: Composition of continuous functions

### Discontinuity of the topological sine function

Math for Non-Geeks: Template:Aufgabe