Sequential definition of continuity – Serlo

Motivation and derivation

First examples

Consider the limit $\lim _{n\to \infty }\exp \left({\tfrac {1}{n}}\right)$. In school, this limit would be calculated as follows:

${\begin{aligned}\lim _{n\to \infty }\exp \left({\frac {1}{n}}\right)&=\exp \left(\lim _{n\to \infty }{\frac {1}{n}}\right)\\[0.5em]&{\color {OliveGreen}\left\downarrow \ \lim _{n\to \infty }{\frac {1}{n}}=0\right.}\\[0.5em]&=\exp(0)\\&=1\end{aligned}}$

Intuitively, this calculation makes sense: if ${\tfrac {1}{n}}\to 0$, then $\exp \left({\tfrac {1}{n}}\right)\to \exp(0)$ should hold. But can we really argument like that? Why should we be allowed to "pull" the limit inside the function brackets? Let us consider another example: the sign function $\operatorname {sgn}(x)$, which is returning the sign of $x$:

$\operatorname {sgn}(x)={\begin{cases}1&x>0\\0&x=0\\-1&x<0\end{cases}}$

Since ${\begin{aligned}{\tfrac {1}{n}}>0\end{aligned}}$ there is:

$\lim _{n\to \infty }\operatorname {sgn} \left({\tfrac {1}{n}}\right)=\lim _{n\to \infty }1=1\neq 0=\operatorname {sgn}(0)=\operatorname {sgn} \left(\lim _{n\to \infty }{\frac {1}{n}}\right)$

So $\lim _{n\to \infty }\operatorname {sgn} \left({\tfrac {1}{n}}\right)\neq \operatorname {sgn} \left(\lim _{n\to \infty }{\tfrac {1}{n}}\right)$. This simple example shows that the limit may not simply be pulled into the function brackets. The function plot shows, why this worked with $\exp(x)$, but not with $\operatorname {sgn}(x)$. For $\exp(x)$, the sequence $\left(\exp \left({\tfrac {1}{n}}\right)\right)_{n\in \mathbb {N} }$ converges to $\exp(0)$, when taking the limit $n\to \infty$:

exp(1/n) converges to exp(0) for n going to infinity

The sign function has a "jump" in the graph at $x=0$ and hence, the sequence $\left(\operatorname {sgn} \left({\tfrac {1}{n}}\right)\right)_{n\in \mathbb {N} }$ does not converge to $\operatorname {sgn}(0)$:

sgn(1/n) does not converge to(0) for n going to infinity

We note: There are certain functions, where the limit may be pulled into the brackets, wile for other functions this may not always work.

Jumps and continuity

The reason why we cannot pull the limit into the sign function $\operatorname {sgn} \left({\tfrac {1}{n}}\right)$, is because its graph has a "jump" at $x=0$. Let us now try to figure out, why pulling the limit into the brackets does not work, if the graph has a jump at the limit of the argument sequence. We assume $f$ to have the following graph:

Graph of a function f with a jump at x0, where the right-handed limit at x0 exists.

When approaching $x_{0}$ from the left, the function values will also approach $f(x_{0})$. So if the sequence of arguments consists only or almost only (meaning: with finitely many exceptions) of real numbers smaller or equal $x_{0}$, then we may pull the limit inside the brackets. However, this fails if infinitely many elements bigger than $x_{0}$ appear in the sequence of arguments. Their function values will not approach $f(x_{0})$, as the graph is having a jump at $x_{0}$, when looking to the right. This jump causes a minimal distance which function values in the proximity and right of $x_{0}$ cannot fall below. The jump at $x_{0}$ prevents us from being able to pull the limit into the brackets for some sequences of arguments.

The same happens if the graph has a jump at $x_{0}$ in the left-handed direction:

Graph of the function f with a jump at x0, where the left-handed limit at x0 exists.

Here, pulling the limit in fails whenever the sequence of arguments contains infinitely many elements smaller than $x_{0}$. Function values left of $x_{0}$ will not approach $f(x_{0})$ due to the jump.

Mind that the situation may be different if $f$ is not defined at the jump:

Graph of a function f with a jump at x0, where the function is not defined.

Here, the expression $f\left(\lim _{n\to \infty }x_{n}\right)$ does not make sense, since $f$ is not defined at $x_{0}$. So we do not need to consider, whether pulling in the limes is allowed there. For all other real numbers, the graph of $f$ is continuous. We observe: A jump does only cause discontinuity of a function $f$, if $f$ is even defined at the position of the jump.

Transition to a formal definition

Let us take again a function $f$ with a jump at $x_{0}$. When approaching the argument $x_{0}$ from one side, a certain minimal distance between $f(x)$ and $f(x_{0})$ will always be kept. This minimal distance between $f(x)$ and $f(x_{0})$ is caused by the jump at $x_{0}$. When approaching $x_{0}$ from the other side, the $f(x)$-values will come arbitrarily close to $f(x_{0})$ (provided there is no second jump).

For $x$-values that should approach arbitrarily (=„infinitely“) close to $x_{0}$, we can use the notion of a sequence. To do so, we treat the $x$-values as a sequence $(x_{n})_{n\in \mathbb {N} }$, converging to $x_{0}$. The notion of a sequence is useful, as we need usually need infinitely many $x$-values for the approach and sequences are containing infinitely many elements as well.

Now, let us assume that we approach $x_{0}$ from the side, where our $f(x_{n})$ obey a minimal distance to $f(x_{0})$ for $x$ in the vicinity of $x_{0}$. This minimal distance will be sustained in the limit process $\lim _{n\to \infty }$. In case that $\lim _{n\to \infty }f(x_{n})$ exists, we can therefore be sure that $\lim _{n\to \infty }f(x_{n})\neq f(x_{0})$.

However, in our example we may also choose the sequence $(x_{n})_{n\in \mathbb {N} }$ such that $\lim _{n\to \infty }f(x_{n})=f(x_{0})$. This will, for instance, be the case when $x_{0}$ is approached by our $x_{n}$ from the other side. So in order to have $\lim _{n\to \infty }f(x_{n})\neq f(x_{0})$, we cannot make an arbitrary choice of our sequence converging to $x_{0}$. But we certainly know that there are sequences $(x_{n})_{n\in \mathbb {N} }$, for which $f(x_{n})$ does not tend towards $f(x_{0})$.

Derivation of the sequence criterion

Let us recap what we found so far:

Whenever a function $f:D\to \mathbb {R}$ has a jump at $x\in D$ , then we may find at least one sequence $(x_{n})_{n\in \mathbb {N} }$ of arguments with $\lim _{n\to \infty }x_{n}=x$ but $\lim _{n\to \infty }f(x_{n})\neq f(x)$.

So if there is a jump at $x\in D$ then we may not pull the limit inside the brackets (i.e. $\lim _{n\to \infty }f(x_{n})\neq f(x)=f\left(\lim _{n\to \infty }x_{n}\right)$) for at least one sequence of arguments $(x_{n})_{n\in \mathbb {N} }$ converging to $x$. Now, our intuition tells us that a function is discontinuous at some argument $x$, whenever its graph has a jump there. Therefore, we may define:

The graph of a function $f:D\to \mathbb {R}$ is discontinuous at some argument $x\in D$, if there is at least one sequence $(x_{n})_{n\in \mathbb {N} }$ of arguments with $\lim _{n\to \infty }x_{n}=x$ but $\lim _{n\to \infty }f(x_{n})\neq f(x)$ .

In order to find the corresponding definition for continuity at $x$, we just need to take the negation of the above definition for discontinuity. That means, we may understand continuity as absence of a jump, where the exact definition reads:

A function $f:D\to \mathbb {R}$ is continuous at $x\in D$ , if for all sequences $(x_{n})_{n\in \mathbb {N} }$ of arguments in $D$ with $\lim _{n\to \infty }x_{n}=x$ we may pull the limit inside the brackets:

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

So we memorize: in case a function is continuous at some argument $x$, we may pull the limit inside the brackets for any sequence of arguments convrging to $x$. This is the sequence criterion of continuity. The entire function $f$ is considered to be continuous, if it is continuous at each argument $x\in D$ in its domain of definition. The definition of continuity for a function $f$ therefore reads:

A function $f:D\to \mathbb {R}$ is continuous, if for all convergent sequences $(x_{n})_{n\in \mathbb {N} }$ , there is $\lim _{n\to \infty }f(x_{n})=f(\lim _{n\to \infty }x_{n})$. Here, all elements $x_{n}$ of the sequence and the limit $\lim _{n\to \infty }x_{n}$ must be inside the domain $D$ of $f$ , since otherwise the equation $\lim _{n\to \infty }f(x_{n})=f\left(\lim _{n\to \infty }x_{n}\right)$ would make no sense.

So let us recap: For continuous functions $f$ , we may always pull the limit inside the brackets – regardless of the limit $x$ of the sequence of arguments. For instance, the exponential function is continuous, so we may always pull the limit inside the brackets of this function. The sign function is discontinuous at $x=0$ , so we may not pull the limit inside the brackets, if the sequence of argument converges to zero.

Formal definition

In the above section, we already learned the definition of continuity. Let us formalize this:

Definition (Sequence criterion of continuity for a single argument)

A function $f:D\to \mathbb {R}$ with $D\subseteq \mathbb {R}$ is continuous at an argument $x_{0}\in D$, if for all squences $(x_{n})_{n\in \mathbb {N} }$ with $\forall n\in \mathbb {N} :x_{n}\in D$ and $\lim _{n\to \infty }x_{n}=x_{0}$ there is:

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

A function is said to be continuous if it is continuous at each argument:

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:

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

So continuity guarantees us that we are allowed to pull the limit inside the brackets. This may simplify calculations of limits tremendously - and indeed, continuity is a concept one may encounter frequently when calculating limits.

Consequences of this definition

We just found a formal definition of continuity by using our intuition of jumps. This definition turned out to be useful over the years and is therefore accepted in the mathematical community as the definition of continuity. And we will use it for the following considerations as well. So if we want to decide whether a function is continuous, we will replace our first intuition by the sequence criterion which we just found. This will also have some side effects. We consider the topological sine function:

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

Its graph is given by:

Graph der topologischen Sinusfunktion

Our first intuition is not suitable for deciding, whether this function is continuous at $x=0$ , since it is oscillating here infinitely many times with the period of oscillation tending to zero. The graph itself does not look as if there would be a jump. However, using the sequence criterion of continuity, we may show that the topological sine function is discontinuous at $x=0$, which corresponds to the infinitely fast oscillations (Exercise).

As a byproduct of our definition, we found a type of discontinuity differing from a jump. It is also called essential discontinuity . And we need to accept these kind of discontinuities, if we want to use the sequence criterion as a definition for discontinuities. Please note that this is one of the standard definitions in mathematics. So denying it or looking for an alternative would certainly make us some lonely outsiders in the world of mathematicians!

Examples for the sequence criterion

Quadratic functions are continuous

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.

Application of the sequence criterion

We have just proven that the quadratic funtion is continuous. So we may pull the limit inside the brackets without further thinking about it. This is a nice thing about continuity, as the following example underlines:

Exercise (Application of the sequence criterion)

Compute $\lim _{n\to \infty }\left(1+{\tfrac {1}{n}}\right)^{2}$.

Solution (Application of the sequence criterion)

We may use continuity of the quadratic function by pulling the limit inside the brackets. This is allowed by the sequence criterion. And it yields us:

${\begin{aligned}&\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{2}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{quadratic function is continuous}}\right.}\\[0.3em]=\ &\left(\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)\right)^{2}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lim _{n\to \infty }\left(a_{n}+b_{n}\right)=\lim _{n\to \infty }a_{n}+\lim _{n\to \infty }b_{n}\right.}\\[0.3em]=\ &\left(\lim _{n\to \infty }1+\lim _{n\to \infty }{\frac {1}{n}}\right)^{2}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lim _{n\to \infty }{\frac {1}{n}}=0\right.}\\[0.3em]=\ &(1+0)^{2}=1\end{aligned}}$

Common sketches of proofs

Proofs of continuity using the sequence criterion

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:

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

Proving discontinuity using the sequence criterion

In order to show that a function $f:D\to \mathbb {R}$ is discontinuous at $x_{0}\in D$ using the sequence criterion, we need to find one specific sequence of arguments $(x_{n})_{n\in \mathbb {N} }$ with $x_{n}\in D$ for all $n\in \mathbb {N}$ which is converging to $x_{0}$ , such that the sequence of function values $\left(f(x_{n})\right)_{n\in \mathbb {N} }$ does not converge to $f(x_{0})$ . So there shall be $\lim _{n\to \infty }x_{n}=x_{0}$ but $\lim _{n\to \infty }f(x_{n})\neq f(x_{0})$ . In order for $\lim _{n\to \infty }f(x_{n})\neq f(x_{0})$ to hold, there are two cases to be distinguished:

• The sequence of function values $\left(f(x_{n})\right)_{n\in \mathbb {N} }$ diverges.
• The sequence of function values $\left(f(x_{n})\right)_{n\in \mathbb {N} }$ converges, but its limit is not $f(x_{0})$.

Therefore, a proof of discontinuity using the sequence criterion could take the following form:

Let $f:\ldots$ be a function defined by $f(x)=\ldots$. This function is discontinuous at $x_{0}=\ldots$ for the following reason: We take the sequence $(x_{n})_{n\in \mathbb {N} }$ with $x_{n}=\ldots$ and all these elements are inside the domain of definition for $f$. This sequence converges to

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

However, there is $\lim _{n\to \infty }f(x_{n})\neq f(x_{0})$. And instead, there is ...Proof that $\left(f(x_{n})\right)_{n\in \mathbb {N} }$ diverges or that the limit of $\left(f(x_{n})\right)_{n\in \mathbb {N} }$ exists and is different from $f(x_{0})$ ...

Equivalence to the epsilon-delta 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:

For every $\epsilon >0$ , there is a $\delta >0$ such that $|f(x)-f(x_{0})|<\epsilon$ for all $x\in D$ with $|x-x_{0}|<\delta$ .

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:

For any $\epsilon >0$ there has to be an $N\in \mathbb {N}$ such that $|f(x_{n})-f(x_{0})|<\epsilon$ for all $n\geq N$.

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:

There is a $\delta >0$, such that $|f(x_{n})-f(x_{0})|<\epsilon$ for all sequence elements $x_{n}$ fulfilling $|x_{n}-x_{0}|<\delta$ .

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

Continuity of the absolute 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.