# How to proof the convergence of functions without knowing the limit

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Today I had the problem to proof the existence of a real valued function limit $$\lim_{x\rightarrow x_0} f(x)$$ but there was no possibility to determine its value. So I thought about the ways how someone can proof that a function converges while there is no chance that he can guess the limit. How can he do this?

For real valued sequences there is an easy answer: If you can show that your sequence is a Cauchy sequence than your sequence converges. This is stated by the completeness axiom of the real numbers and is true for all complete metric spaces. But there is no concept to generalize the idea of Cauchy sequences to functions. Really?!

## The Cauchy property for functions

Let's have a look at the epsilon-delta-definition for Cauchy sequences in $$\R$$ or $$\C$$:

$\forall \epsilon > 0\, \exists N \in \N\, \forall n,m \in \N: n \ge N \land m \ge N \Rightarrow |a_n-a_m| < \epsilon$

I guessed what would be the equivalent definition for a function, if someone want to proof the existence of $$\lim_{x\rightarrow x_0} f(x)$$ for a real valued function $$f : D\rightarrow \R$$ with $$D \subseteq \R$$:

$\forall \epsilon > 0\, \exists \delta > 0\, \forall x,y\in D: |x-x_0| < \delta \land |y-x_0|<\delta \Rightarrow |f(x)-f(y)| < \epsilon$

Because I want you to understand my thoughts I show you both statements where I colored for me equivalent parts in the same color:

$\begin{array}{lllclcl} \color{royalblue}{\forall \epsilon > 0} & \color{gray}{\exists N \in \N} & \color{magenta}{\forall n,m \in \N} & : & \color{green}{n \ge N \land m \ge N }& \Rightarrow & \color{maroon}{|a_n-a_m| < \epsilon} \\ \color{royalblue}{\forall \epsilon > 0} & \color{gray}{\exists \delta > 0} & \color{magenta}{\forall x,y\in D} & : & \color{green}{|x-x_0| < \delta \land |y-x_0|<\delta} & \Rightarrow & \color{maroon}{|f(x)-f(y)| < \epsilon} \end{array}$

You see, I got the above statement for functions by translating each part of the epsilon-delta-definition of Cauchy sequences in the concept of function or (better to say) I assumed the right translations. Okay, now the main question is: Does the function limit exists, if the above statement is fulfilled?

Before answering the question, lets give functions with the above property a name:

Definition (functions with the Cauchy property at $$x_0$$): Let $$f: D\rightarrow$$ with $$D \subseteq \R$$ be a real valued function and $$x_0$$ an limit point of $$D$$. In the following we say, that $$f$$ has the Cauchy property at $$x_0$$, if $$f$$ fulfills the following statement at $$x_0$$:

$\forall \epsilon > 0\, \exists \delta > 0\, \forall x,y\in D: |x-x_0| < \delta \land |y-x_0|<\delta \Rightarrow |f(x)-f(y)| < \epsilon$

### Every function with the Cauchy property at $$x_0$$ converges in the limit $$x \rightarrow x_0$$

Our interesting question is: Does the limit $$\lim_{x\rightarrow x_0} f(x)$$ exists when $$f$$ has the Cauchy property at $$x_0$$? Yes, as we can proof:

Step 1: Let $$(x_n)_{n\in\N}$$ be an arbitrary sequence with $$\lim_{n\rightarrow\infty} x_n = x_0$$. We can show, that $$\left( f(x_n) \right)_{n\in\N}$$ is a Cauchy sequence and therefore converges.

Let $$\epsilon > 0$$ be arbitrary. By the Cauchy property of $$f$$ at $$x_0$$ there exists a $$\delta > 0$$ so that $$| f(x) - f(y)| < \epsilon$$ for all $$x,y\in D$$ with $$|x-x_0| < \delta$$ and $$|y-x_0| < \delta$$. Because $$(x_n)_{n\in\N}$$ converges to $$x_0$$ there is an $$N \in \N$$ with $$|x_n-x_0| < \delta$$ for all natural numbers $$n \ge N$$.

So for all $$n,m \in \N$$ with $$n \ge N$$ and $$m \ge N$$ we have $$|x_n-x_0| < \delta$$ and $$|x_m-x_0| < \delta$$ and therefore $$|f(x_n)-f(x_m)| < \epsilon$$ because of the above definition of $$\delta$$. This shows that $$\left(f(x_n)\right)_{n\in\N}$$ is a Cauchy sequence. So there is an $$a\in\R$$ with $$\lim_{n\rightarrow\infty} f(x_n) = a$$ because of the completeness axiom of the real numbers.

Step 2: Now we have to show that every two sequences $$(x_n)_{n\in\N}$$ and $$(\tilde x_n)_{n\in\N}$$ which converge to $$x_0$$ have the same limit under the function $$f$$. This means $$\lim_{n\rightarrow \infty} f(x_n) = \lim_{n\rightarrow \infty} f(\tilde x_n)$$.

In Step 1 we already showed that the limit $$\lim_{n\rightarrow \infty} f(x_n)$$ exists. So let the limit of $$(f(x_n))_{n\in\N}$$ be $$a \in \R$$. Let $$\epsilon > 0$$ be arbitrary. Because $$f$$ has the Cauchy property at $$x_0$$ there exists a $$\delta > 0$$ so that $$| f(x) - f(y)| < \tfrac 12 \epsilon$$ for all $$x,y\in D$$ with $$|x-x_0| < \delta$$ and $$|y-x_0| < \delta$$. Because $$\lim_{n\rightarrow\infty} x_n = x_0$$ and $$\lim_{n\rightarrow\infty} f(x_n) = a$$ there is a $$N \in \N$$ with $$|x_N -x_0| < \delta$$ and $$|f(x_N)-a| < \tfrac 12 \epsilon$$. There is also an $$\tilde N \in \N$$ with $$|\tilde x_n-x_0| < \delta$$ for all $$n\in \N$$ with $$n \ge \tilde N$$.

Now let $$n$$ be an arbitrary natural number greater than $$\tilde N$$. By definition we have $$|\tilde x_n-x_0| < \delta$$ and $$|x_N-x_0| < \delta$$ and therefore $$|f(\tilde x_n)-f(x_N)| < \tfrac 12 \epsilon$$. So we have:

$|f(\tilde x_n)-a| \le |f(\tilde x_n)-f(x_N)| + |f(x_N)-a| < \tfrac 12 \epsilon + \tfrac 12 \epsilon = \epsilon$

This proofs $$\lim_{n\rightarrow \infty} f(\tilde x_n) = a$$ and thus $$\lim_{n\rightarrow \infty} f(x_n) = \lim_{n\rightarrow \infty} f(\tilde x_n)$$. So the function limit $$\lim_{x\rightarrow x_0} f(x)$$ exists.

### If $$\lim_{x\rightarrow x_0} f(x)$$ exists, then $$f$$ has the Cauchy property at $$x_0$$

Also the contrary implication is true: Every function $$f$$, which converges at $$x_0$$, has the Cauchy property at $$x_0$$.

Proof: Let $$f$$ be a function with $$\lim_{x\rightarrow x_0} f(x) = a$$ an let $$\epsilon$$ be an arbitrary real number greater than null. Because $$f$$ converges to $$a$$ for $$x \rightarrow x_0$$ there is an $$\delta > 0$$ so that the inequality $$|f(x)-a| < \tfrac 12 \epsilon$$ is fulfilled for every $$x \in D$$ with $$|x-x_0|<\delta$$. Now for every $$x,y \in D$$ with $$|x-x_0| < \delta$$ and $$|y-x_0| < \delta$$ we have:

$|f(x)-f(y)| \le |f(x)-a| + |f(y)-a| < \tfrac 12 \epsilon + \tfrac 12 \epsilon = \epsilon$

### Connection to the completeness axiom

We have shown, that the following equivalence statement is true:

$\begin{array}{c} \lim_{x \rightarrow x_0} f(x) \text{ exists} \\ \Leftrightarrow \\ f \text { has the Cauchy property at } x_0 \end{array}$

Does this remind you an something? Yes, it's similar to the completeness axiom! In the article Alternative formulations of the completeness axiom for real and complex numbers I will proof that the above equivalence actually is equivalent to the completeness axiom.

### Some additional notes to functions with the Cauchy property at $$x_0$$

• Please note, that $$x_0$$ does not have to be an element of $$D$$. There just have to exist a sequence in $$D$$ which converges to $$x_0$$.

• Because we are interested in proofing the existence of $$\lim_{x\rightarrow x_0} f(x)$$ at a special point $$x_0$$ the above definition for the Cauchy property of a function also depends on selecting a special $$x_0$$. It would be nice to know, what a global Cauchy property for functions might be, but I do not want to provide an answer in this article. Please contact me if you have an answer.

• The above definition can be directly generalized to complex functions $$f: D\rightarrow \C$$ with $$D \subseteq \C$$. All the proofs given in this article can also be applied to this case.

• There shall be a more general definition for functions between metric spaces. Because metric spaces do not have those properties which are needed in the given proofs, those proofs cannot be applied directly to functions between metric spaces. One has to find the necessary properties and has to take care that the underlying metric spaces have fulfilled those properties.

• There is also a Cauchy property of $$f$$, if someone wants to show the convergence for $$x \rightarrow \infty$$:

$\forall \epsilon > 0\, \exists S > 0\, \forall x,y\in D: x\ge S \land y\ge S \Rightarrow |f(x)-f(y)|<\epsilon$

For showing the existence of $$\lim_{x\rightarrow -\infty} f(x)$$ the definition would be:

$\forall \epsilon > 0\, \exists S < 0\, \forall x,y\in D: x\le S \land y\le S \Rightarrow |f(x)-f(y)|<\epsilon$

To have a short article I do not want to provide proofs for the above definitions but I assume that they will be similar.

## Alternative ways to proof the convergence of a function without knowing the limit

It is nice to have as many possible ways or methods because than it is more likely you find actually a proof than if you stuck by going just one way. So I want to enumerate also the other possibilities for proofing the existence of $$\lim_{x\rightarrow x_0} f(x)$$:

One alternative way is of course to use an indirect proof, which means that you assume the function is divergent at $$x_0$$ and that you conclude a contradiction from your assumption. There just might be the situation where you do want to have a direct proof. For example you are doing math based on intuitionistic logic or you want to know the insights which just would be provided by a direct proof.

Another way would be to give an algorithm to calculate the limit and to show that this algorithm will always work. Here you have the advantage that you can use the algorithm in a computer program. On the contrary it might be difficult to find such an algorithm.

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• # How to compare infinite sets of natural numbers, so that proper subsets are also strictly smaller than their supersets

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• # Alternative formulations of the completeness axiom for real and complex numbers

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