Convergent Functions
We have seen convergent sequences where as the index \(n\) of the sequence approaches infinity, the sequence converges to a limit \(L\). We have also seen that this concept of convergence exists for series, allowing us to define convergent series. In this section we will look at the concept of convergence for functions.
Accumulation Points
Before we can define the limit of a function, we need to understand the concept of accumulation points or also often called limit points. An accumulation point of a set \(D\) is a point \(x_0\) such that every neighborhood of \(x_0\) contains at least one point of \(D\) different from \(x_0\), more formally \(x_0\) is an accumulation point of \(D\) if the following holds:
\[\forall \epsilon > 0 \quad ((x_0 - \epsilon, x_0 + \epsilon) \setminus \{x_0\}) \cap D \neq \emptyset \]If we define \(D=\{0\} \cup (1,2)\), then \(x_0=0\) is not an accumulation point of \(D\) since we can take the neighborhood where \(\epsilon=1\) and we have \((0-1,0+1) \setminus \{0\}\) which is \((-1,1)\) except for the point \(0\) and this neighborhood does not contain any point of \(D\). They nearly touch at the point \(1\) but don’t. The same would hold for \(\epsilon=0.5\) and so on so \(0\) is not an accumulation point of \(D\).
However, if we take \(x_0=1\), then for any \(\epsilon>0\) we have \((1-\epsilon,1+\epsilon) \setminus \{1\}\) which is guaranteed to contain some point larger than \(1\). Because our contains the interval \((1,2)\), there is also a point larger than \(1\) in the interval \((1,2)\) and hence \(1\) is an accumulation point of \(D\).
The reverse is also true, if we take \(x_0=2\), then for any \(\epsilon>0\) we have \((2-\epsilon,2+\epsilon) \setminus \{2\}\) which is guaranteed to contain some point smaller than \(2\) and again because our set contains the interval \((1,2)\), there is also a point smaller than \(2\) in the interval \((1,2)\) and hence \(2\) is an accumulation point of \(D\).
This is actually the case for any point in the interval \([1,2]\) and hence the set of all accumulation points of \(D\) denoted as \(D'\) is \(D'=[1,2]\).
Another example would be if we take the set \(D=(1,2) \cup (3,5)\). Then the set of all accumulation points of \(D\) is \(D'=[1,2] \cup [3,5]\). For the points between 2 and 3, so \(x \in (2,3)\), we can show that they are not accumulation points. Because if we pick any \(\epsilon > 0\) less or equal to the following value we don’t have any points in the neighborhood of \(x\) that are also in \(D\):
\[\epsilon = \frac{\min(3-x,x-2)}{2} \]So for example if we take \(x=2.2\) we take the minimum of \(3-2.2=0.8\) and \(2.2-2=0.2\) which is \(0.2\) and we can take \(\epsilon=0.1\). So we have \((2.2-0.1,2.2+0.1) \setminus \{2.2\}\) which is \((2.1,2.3)\) and this neighborhood does not contain any point of \(D\). So \(x=2.2\) is not an accumulation point of \(D\). The same would hold for any point in the interval \((2,3)\). To illustrate this we could take \(x=2.001\). The minimum of \(3-2.001=0.999\) and \(2.001-2=0.001\) is \(0.001\) and we can take \(\epsilon=0.0005\) to then have \((2.001-0.0005,2.001+0.0005) \setminus \{2.001\}\) which is \((2.0005,2.0015)\) and this neighborhood does not contain any point of \(D\). So \(x=2.001\) is not an accumulation point of \(D\).
Limit of a Function
Now that we understand the concept of accumulation points, we can define the limit of a function. First we define the function \(f\) on a set \(D \subseteq \RR\) as a mapping from \(D\) to \(\RR\):
\[f:D \to \RR \]if \(x_0 \RR\) is then an accumulation point of \(D\) then we write that the function \(f\) has a limit \(L\) at the point \(x_0\) as follows:
\[\lim_{x \to x_0}{f(x)}=L \]if for every \(\epsilon > 0\) there exists a \(\delta > 0\) such that the following holds:
\[\forall \epsilon > 0 \quad \exists \delta > 0 \quad \forall x \in ((x_0 - \delta, x_0 + \delta) \setminus \{x_0\}) \cap D \quad |f(x) - L| < \epsilon \]Because we know that \(x_0\) is an accumulation point of \(D\) we know that there exists at least one point \(x\) in the neighborhood of \(x_0\) that is also in \(D\). We can interpret the definition of the limit of a function similarly to the epsilon-delta definition of continuity. As we approach the point \(x_0\) from both sides, so making \(\delta\) smaller and smaller, the function \(f(x)\) approaches the limit \(L\). Or in other words, for any \(\epsilon\)-neighborhood of \(L\) we can find a point \(x\) in the neighborhood of \(x_0\) such that \(f(x)\) is in the \(\epsilon\)-neighborhood of \(L\). The requirement for the point \(x_0\) to be an accumulation point of \(D\) is important because if it is not, then we may not be able to find a point \(x\) where \(f(x)\) is in the \(\epsilon\)-neighborhood of \(L\).
Another way of defining the limit of a function is to use sequences. If \(x_0\) is an accumulation point of \(D\) and we take any sequence \(a_n\) in \(D \setminus \{x_0\}\) such that the sequence converges to \(x_0\) as \(n\) approaches infinity, so \(\lim_{n \to \infty} a_n = x_0\), then we can say that the limit of the function \(f\) at the point \(x_0\) is equal to the limit of the sequence \(a_n\) as \(n\) approaches infinity:
\[\lim_{n \to \infty} f(a_n) = L \]The idea is that as the sequence \(a_n\) approaches the point \(x_0\), the function \(f(a_n)\) approaches the limit \(L\). So in a way we are again saying that the function \(f\) approaches the limit \(L\) as we approach the point \(x_0\).
We are given the function \(f(x) = 2x\) and we want to inspect the limit at the point \(x_0=0\). We can take the sequence \(a_n = \frac{1}{n}\) which converges to \(0\) as \(n\) approaches infinity. We can then calculate the limit of the function at the point \(x_0=0\) by substituting the sequence into the function:
\[\lim_{x \to 0} f(x) = \lim_{n \to \infty} f(a_n) = \lim_{n \to \infty} f(\frac{1}{n}) = \lim_{n \to \infty} 2(\frac{1}{n}) = 0 \]So we can say that the limit of the function \(f\) at the point \(x_0=0\) is \(0\). We can also take the sequence \(a_n = -\frac{1}{n}\) which also converges to \(0\) as \(n\) approaches infinity. Again applying the sequence to the function we get:
\[\lim_{x \to 0} f(x) = \lim_{n \to \infty} f(a_n) = \lim_{n \to \infty} f(-\frac{1}{n}) = \lim_{n \to \infty} 2(-\frac{1}{n}) = 0 \]Limit of a Function at Infinity
Wir können den Funktionswert an der Stelle \(x = 0\)zwar nicht berechnen, aber mit einer Folge \(x_n\) beliebig nahe an die Definitionslücke herantasten.
What about accumulation points at the point \(x_0=\infty\)? or if we have a function that is not defined at \(x_0\)?
Left and Right Limits
Zusammenfassung Grenzwert einer Funktion
Betrachtet man bei der Grenzwertbetrachtung einer Funktion \(f\) an der Stelle \(x_0\) nur Zahlenfolgen \(x_n\), die kleinere Werte als \(x_0\) enthalten, dann bezeichnet man den Grenzwert als Linkseitigen Grenzwert
\[\lim_{x \to x_0,(x<x_0)}{f(x)}=\lim_{h \to 0,(h>0)}{f(x_0-h)}=G_L \]Zahlenfolgen mit grösseren Werten als \(x_0\) erzeugen den Rechtseitigen Grenzwert
\[\lim_{x \to x_0,(x>x_0)}{f(x)}=\lim_{h \to 0,(h>0)}{f(x_0+h)}=G_R \]Streben für jede gegen \(x_0\) konvergente Zahlenfolge \(x_n\) die Funktionswerte \(f(x_n)\) gegen denselben Wert \(G\), dann besitzt die Funktion \(f\) an der Stelle \(x_0\) den Grenzwert \(G\)
\[\lim_{x \to x_0}{f(x)}=G=G_R=G_L \]Rechenregeln für Funktionsgrenzwerte
Seien \(f\) und \(g\) zwei Funktionen mit dem gleichen Grenzwerten \(G,F\) bei \(x_0\) dann gilt:
- \(\lim_{x \to x_0}{(f(x)\pm g(x))}=F\pm G\)
- \(\lim_{x \to x_0}{(f(x)*g(x))}=F*G\)
- \(\lim_{x \to x_0}{(f(x)\over g(x))}=F\over G\) für \(g(x_0) \neq 0\) und \(G \neq 0\)