(last updated: 2:32:29 PM, October 30, 2020)
Definition 3.2.1 Suppose that a function \(f : D \rightarrow \mathbb R\), with \(D\) a subset of \(\mathbb R\), and suppose that \(a\) is an accumulation point of \(D\). The function \(f\) has a limit as \(x\) approaches (or as \(x\) tends to) \(a\) if and only if there exists a real number \(L\) such that for every \(\epsilon > 0\) there exists a real number \(\delta > 0\) such that \(| f (x) - L |< \epsilon\) provided that \(0< |x - a| < \delta\) and \(x \in D\). Whenever this is the case, we write \(\displaystyle\lim_{x\rightarrow a}f (x) = L\).
Note that the limiting value here can be any real number. THe case where the funciton goes to \(\pm\infty\) is covered later.
Example:
Let \(f(x)=\frac{x}{5}\). Prove that \(f(x)\overset{x\rightarrow 5}{\longrightarrow} 1\).
Proof. The domain is all real numbers since no specific restrictions were stated. This is often referred to as the natural domain of the function. Let \(\epsilon>0\) be some arbitrary positive number, and let \(\delta=5\epsilon\). Then when \(0<|x-5|<\delta\) we have that \(|f(x)-1|=|\frac x5-1|=\frac{|x-5|}{5}<\frac{\delta}{5}=\epsilon\). \(\square\)
Note that for this definition to make sense, the point \(a\) must be an accumulation point of the domain \(D\).
Example & Counterexample:
Let \(f(x)=x^2+2\) for \(x\in\{\frac1n \mid n\in\mathbb N\}=D\). This is a function whose natural domain is \(\mathbb R\), but we have restricted its domain. This domain restriction is part of the functions definition, i.e. \(f(0)\) is undefined; it does not exist.
Note that \(0\) is an accumulation point of the domain though (but \(x=0\) is not allowed since \(0\not\in D\)). Also note that \(1\) is not an accumulation point (and \(x=1\) is allowed since \(1\in D\)).
Therefor \(\displaystyle \lim_{x\rightarrow0}f(x)=2\) exists, but \(\displaystyle \lim_{x\rightarrow1}f(x)\) does not exist, since we are unable to evaluate \(f(x)\) “close” to but not at \(x=1\).
Proof. For any \(\epsilon>0\), choose \(\delta=\sqrt{\epsilon}\). Then \(0<x<\delta\) with \(x\in D\) will give us \(|f(x)-2|=|(x^2+2)-2|=x^2<(\sqrt\epsilon)^2=\epsilon\). Note that we didn’t really use the strange form of the domain \(D\) except to note that there are always infinitely many \(x\in D\) such that \(0<x<\delta\) for any positive \(\delta\).
Now we might feel like \(f(x)\rightarrow 3\) as \(x\rightarrow 1\), but due to our strange domain, this isn’t the case. Note that \(f(1)=3\) and that the closest value in the domain to \(x=1\) is \(x=\frac12\) but \(f(\frac12)=\frac14+2=2.25\). Let \(\epsilon=0.5\). Then we want to find an \(x\in D\) with \(x\not= 1\) such that \(|f(x)-3|<0.1\). However, this means we want \(2.9<f(x)<3.1\), but that is only true for \(x=1\) in the domain. Thus \(\displaystyle \lim_{x\rightarrow1}f(x)\) does not exist. \(\square\)
Theorem 3.2.5. Suppose that functions \(f, g, h : D \rightarrow \mathbb R\), with \(D \subset\mathbb R\), \(a\) is an accumulation point of \(D\), and \(\displaystyle \lim_{x\rightarrow a}f (x) = A\), \(\displaystyle \lim_{x\rightarrow a}g (x) = B\), and \(\displaystyle \lim_{x\rightarrow a}h (x) = C\). Then all of the conclusions for Theorem 3.1.7 are true with \(\infty\) replaced by \(a\) and with “eventually” replaced by “near \(x = a\).”
This is a version of the same result from the previous section.
Definition 3.2.12. Suppose that the function \(f : D \rightarrow\mathbb R\) with \(D\) a subset of \(\mathbb R\) and \(a\) an accumulation point of \(D\). Then the function \(f\) tends to plus infinity as \(x\) approaches, tends to, \(a\) if and only if for any given real number \(K > 0\), there exists \(\delta > 0\) such that \(f (x) > K\), provided that \(0 < |x - a| < \delta\) and \(x \in D\). Whenever this is the case, we write \(\displaystyle\lim_{x\rightarrow a} f (x) = +\infty\).
The above definition can also be modified to define \(f(x)\overset{x\rightarrow a}{\longrightarrow} -\infty\) as well.
Example:
Consider \(f(x)=\frac{1}{(x-2)^2}\). Show that \(f(x)\overset{x\rightarrow 2}{\longrightarrow} \infty\).
Let \(K>0\) be an arbitrary (large) number. Set \(\delta=\frac{1}{\sqrt K}\). Then \(0<|x-2|<\delta\) gives \(f(x)=\frac{1}{(x-2)^2}=\frac{1}{|x-2|^2}>\frac{1}{\delta^2}=\frac{1}{(\frac{1}{\sqrt K})^2}=K\). \(\square\)
The following function is called Thomae’s function, and goes by other names as well, such as the modified Dirichlet function.
Example: Consider the function \[ f(x)=\begin{cases} \frac1q &\text{ for } x=\frac pq\in\mathbb Q \text{ in reduced form}, p\in\mathbb Z, q\in\mathbb N\\ 0 &\text{ otherwise} \end{cases} \] This function satisfies \(\displaystyle\lim_{x\rightarrow a} f(x)=0\) for any \(a\in\mathbb R\).
Proof. Let \(\epsilon>0\) and \(x=\frac pq\in\mathbb Q\) in reduced form with \(q\in\mathbb N\). Pick \(k\in\mathbb N\) such that \(\frac1{kq}<\epsilon\). Now consider the interval \((\frac pq-1,\frac pq+1)\) and all rational numbers in that interval with reduced form \(\frac ab\) where \(b\leq kq\). Notice that \(f(\frac ab)=\frac1b \geq \frac 1{kq} \geq \epsilon\). Let \(S=\{\frac ab \mid \frac ab\in (\frac pq-1,\frac pq+1),b\leq kq\}\). We have that \(\frac pq\in S\) and that any rational number in \((\frac pq-1,\frac pq+1)\) but not in \(S\) will have a denominator greater than \(kq\). Let \(\delta=\min\{|x-y| \mid x,y\in S\}\). Notice that all rational numbers in \((\frac pq-\delta,\frac pq+\delta)\) have denominator greater than \(kq\). Thus \(0<|x-\frac pq|<\delta\) and \(x=\frac cd\in\mathbb Q\) implies that \(f(x)=\frac1d<\frac1{kq}<\epsilon\). Any irrational \(x\in (\frac pq-\delta,\frac pq+\delta)\) of course satisfies \(f(x)=0<\epsilon\). Therefore \(\displaystyle\lim_{x\rightarrow \frac pq} f(x)=0\).
Now let \(a\) be any irrational number. Essentially the same construction works. Let \(S=\{\frac pq\in\mathbb Q\cap (a-1,a+1) \mid \frac pq\text{ is reduced form}, \frac1q\geq\epsilon\}\). Let \(\delta=\min\{|x-y| \mid x,y\in S\cup\{a\}\}\). Then we know that any rational numbers in \((a-\delta,a+\delta)\) will thus not be in \(S\) and therefor have reduced forms like \(\frac cd\) with \(\frac1d<\epsilon\). So that \(0<|x-a|<\delta\) with \(x=\frac cd\) rational satisfies \(f(x)=\frac1d<\epsilon\), and of course \(0<|x-a|<\delta\) and \(x\) irrational satisfies \(f(x)=0<\epsilon\). Therefore we conclude that \(\displaystyle\lim_{x\rightarrow a} f(x)=0\).
We have show that \(\displaystyle\lim_{x\rightarrow a} f(x)=0\) for all rational and irrational \(a\) and thus it is true for all \(a\in\mathbb R\). \(\square\)
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