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Math 382-0: Kursobjekt

Aaron Greicius

Section 1.8 Limits of functions and continuity

Single-variable calculus can be succinctly described as the science of functions of the form \(f\colon D\rightarrow \R\text{,}\) where \(D\subseteq \R\text{.}\) The main tools of that science are the derivative and the integral, both of which rely on the more fundamental concept of the limit.

Subsection Limits of functions

Definition 1.8.1. Limit point.

Given a subset \(A\subseteq \C\text{,}\) a limit point (or accumulation point) of \(A\) is an element \(z_0\in \C\) satisfying the following property: given any \(\epsilon > 0\text{,}\) the open disc \(B_\epsilon(z_0)\) contains an element of \(A\) that is distinct from \(z\text{.}\)

Example 1.8.2. Limit points.

  1. Show that all elements of an open set \(U\) are limit points of \(U\text{.}\)
  2. Given a subset \(Y\subseteq \C\) Show that any point \(z_0\in \partial Y-Y\) is a limit point of \(Y\text{.}\)
  3. Give an example of a set \(Y\) and a boundary point \(z_0\in \partial Y\) that is not a limit point of \(Y\text{.}\)
Solution.

Definition 1.8.3. Limit of a function.

Let \(f\colon D\rightarrow \C\) be a complex function, and let \(z_0\) be a limit point of \(D\text{.}\) We say the limit of \(f\) as \(z\) approaches \(z_0\) exists if there is a complex number \(w\in \C\) satisfying the following property: for all \(\epsilon > 0\text{,}\) there exists a \(\delta> 0\) such that if \(0< \abs{z-z_0} < \delta\text{,}\) then \(\abs{f(z)-w}< \epsilon\text{.}\)
When this is the case we call \(w\) the limit of \(f\) as \(z\) approaches \(z_0\text{,}\) and write \(\lim\limits_{z\to z_0}f(z)=w\text{.}\) Otherwise we say that \(\lim\limits_{z\to z_0}f(z)\) does not exist.

Proof.

We prove each implication separately.
Implication: (1)\(\implies\)(2).
Assume \(\lim\limits_{z\to z_0}f(z)=w\text{,}\) and suppose \(z_n\to z_0\) for a sequence \((z_n)\) consisting of elements of \(D\text{.}\) We show that \(f(z_n)\to w\text{.}\) To this end, given any \(\epsilon> 0\text{,}\) there is a \(\delta> 0\) such that
\begin{equation*} 0< \abs{z-z_0}< \delta \implies \abs{f(z)-w}< \epsilon\text{.} \end{equation*}
Since \(z_n\to z_0\text{,}\) there is an \(N\) such that \(\abs{z-z_n}< \delta\) for all \(n\geq N\text{.}\) It follows that for all \(n\geq N\text{,}\) either \(z_n=z_0\) or \(0< \abs{z-z_n} < \delta\text{.}\) In either case, the displayed implication above implies that \(\abs{f(z_n)-w}< \epsilon\text{,}\) as desired.
Implication: (2)\(\implies\)(1).
We will prove the contrapositive. Suppose that \(\lim\limits_{z\to z_0}f(z)\ne w\text{.}\) By definition, this means there exists an \(\epsilon > 0\) such that for each \(\delta_n=\frac{1}{n}\) there exists a \(z_n\in D\) satisfying \(0< \abs{z_n-z_0}< \frac{1}{n}\) but \(\abs{f(z_n)-w}> \epsilon\text{.}\) It follows easily that \((z_n)\) is a sequence of elements of \(D\) satisfying \(z_n\to z_0\) and \(f(z_n)\not\to w\text{,}\) as desired.

Example 1.8.5. No limit.

Define \(f\colon \C-\{0\}\rightarrow \C\) as \(f(z)=\frac{\overline{z}}{z}\text{.}\) Show that \(\lim\limits_{z\to 0}f(z)\) does not exist.
Solution.
It is easy to see that if the limit existed, then we would have \(\lim\limits_{n\to \infty}f(z_n)=\lim\limits_{n\to \infty}f(w_n)\) for any two sequences \(z_n\to 0\) and \(w_n\to 0\text{.}\) Taking the sequences \((z_n)=\frac{1}{n}\) and \((w_n)=\frac{i}{n}\text{,}\) however, it is easy to see that \(z_n\to 0\) and \(w_n\to 0\text{,}\) and yet
\begin{align*} \lim\limits_{n\to \infty}f(z_n) \amp = \lim\limits_{n\to \infty}\frac{\overline{1/n}}{1/n}=\lim\limits_{n\to \infty}1=1\\ \lim\limits_{n\to \infty}f(w_n) \amp = \lim\limits_{n\to \infty}\frac{\overline{i/n}}{i/n}=\lim\limits_{n\to \infty}-1=-1\text{.} \end{align*}
We conclude that the limit \(\lim\limits_{z\to 0}f(z)\) does not exist.

Proof.

Example 1.8.7. Limit of power function.

Let \(f(z)=z^n\text{,}\) where \(n\) is a positive integer. Prove: for all \(z_0\in \C\text{,}\) \(\lim\limits_{z\to z_0}z^n=z_0^n\text{.}\)
Solution.
There are a number ways of proving this statement. We give a proof by induction for \(n\geq 1\text{.}\)
Base step: \(n=1\text{.}\) Given any \(\epsilon > 0\text{,}\) setting \(\delta=\epsilon\text{,}\) we have
\begin{align*} \abs{z-z_0}< \delta\amp \implies \abs{z-z_0}< \epsilon\\ \amp \implies \abs{f(z)-f(z_0)}< \epsilon\text{,} \end{align*}
since \(f(z)=z^1=z\) and \(f(z_0)=z_0^1=z_0\) in this case. This proves \(\lim\limits_{z\to z_0}f(z)=f(z_0)\) when \(f(z)=z^1\text{.}\)
Induction step: we assume that \(\lim\limits_{z\to z_0}z^n=z_0^n\) and show that \(\lim\limits_{z\to z_0}z^{n+1}=z_0^{n+1}\text{.}\) To do so we use the induction hypothesis and the multiplicative property of limits from Theorem 1.5.5:
\begin{align*} \lim\limits_{z\to z_0}z^{n+1} \amp = \lim\limits_{z\to z_0}z\cdot z^n \\ \amp =\lim\limits_{z\to z_0}z\, \lim\limits_{z\to z_0}z^n \amp \text{(mult. prop.)}\\ \amp = z_0\, z_0^n \amp (\text{base case, ind. hyp.}) \\ \amp =z_0^{n+1} \end{align*}

Subsection Continuity

We end with the notion of continuity. You will notice a strong resemblance between the definition of continuity of a function at a point and the limit of that function at a point. The true nature of this connection is articulated by Theorem 1.8.9.

Definition 1.8.8. Continuous functions.

Let \(f\colon D\rightarrow \C\) be a complex function. Given an element \(z_0\in D\text{,}\) we say that \(f\) is continuous at \(z_0\) if the following property holds: for all \(\epsilon> 0\text{,}\) there exists a \(\delta> 0\) such that if \(\abs{z-z_0}< \delta\text{,}\) then \(\abs{f(z)-f(z_0)}< \epsilon\text{.}\)
We say \(f\) is continuous on a subset \(A\subseteq D\), if \(f\) is continuous at all points \(z_0\in A\text{.}\) We say \(f\) is continuous if \(f\) is continuous on \(D\text{.}\)

Proof.

We will prove the cycle of implications (1)\(\implies\)(2)\(\implies\)(3)\(\implies\)(1).
Implication: (1)\(\implies\) (2).
Assume \(z_0\in D\) is an accumulation point of \(D\) and that \(f\) is continuous at \(z_0\text{.}\) We will show that \(\lim\limits_{z\to z_0}f(z)=f(z_0)\text{.}\) Indeed, since \(f\) is continuous at \(z_0\) given any \(\epsilon > 0\text{,}\) there is a \(\delta> 0\) such that
\begin{equation*} \abs{z-z_0} < \delta \implies \abs{f(z)-f(z_0)}< \epsilon \end{equation*}
for all \(z\in D\text{.}\) But then it is certainly true that
\begin{equation*} 0< \abs{z-z_0} < \delta \implies \abs{f(z)-f(z_0)}< \epsilon \end{equation*}
for all \(z\in D\text{.}\) Thus \(\lim\limits_{z\to z_0}f(z)=f(z_0)\text{.}\)
Implication: (2)\(\implies\)(3).
Assume (2) holds. Given any \(z_0\in D\text{,}\) we wish to show that for all sequences \((z_n)\) of elements of \(D\) satisfying \(z_n\to z_0\text{,}\) we have \(f(z_n)\to f(z_0)\text{.}\) If \(z_0\) is a limit point of \(D\text{,}\) then this follows immediately from Theorem 1.8.4. On the other hand, if \(z_0\) is not a limit point of \(D\text{,}\) it is easy to see that there is no sequence \((z_n)\) of elements of \(D\) satisfying \(z_n\to z_0\text{,}\) and so the condition is satisfied vacuously!
Implication: (3)\(\implies\)(1).
We prove the contrapositive, using an argument very similar to the one in Theorem 1.8.4. Namely, assuming \(f\) is not continuous at \(z_0\in D\text{,}\) we can find an \(\epsilon> 0\) and a sequence \((z_n)\) of elements of \(D\) satisfying \(\abs{z_n-z_0}< \frac{1}{n}\) and \(\abs{f(z_n)-f(z_0)}> \epsilon\text{.}\) It follows that \(z_n\to z_0\) but \(f(z_n)\not\to f(z_0)\text{,}\) as desired.

Proof.

We content ourselves to prove this statement for a few examples.
Consider \(f(z)=e^z\text{.}\) For \(z=x+iy\) we have \(f(z)=u(x,y)+iv(x,y)\text{,}\) where
\begin{align*} u(x,y) \amp = e^{\sqrt{x^2+y^2}}\cos y \amp v(x,y)\amp =e^{\sqrt{x^2+y^2}}\sin y\text{.} \end{align*}
Using continuity properties of real-valued functions of two variables, we see easily that \(u\) and \(v\) are both continuous everywhere. Thus \(f(z)=e^z\) is continuous. (It then follows easily that all of the trigonometric and hyperbolic functions are continuous, since they are defined in terms of \(f(z)=e^z\) using elementary operations.)
Consider \(\Log=u+iv\text{,}\) where \(u(z)=\ln\abs{z}\) and \(v(z)=\Arg z\) for all \(z\in \C-(-\infty, 0]\text{.}\) Since the identity function \(f(z)=z\) is obviously continuous, so is \(\abs{f}=\abs{z}\text{.}\) Since \(\ln\colon (0,\infty)\rightarrow \R\) is continuous, so is the composition \(ln\abs{z}\text{.}\) Lastly, you showed in your homework that \(\Arg\colon \C-(-\infty, 0]\rightarrow \C\) is continuous. Since the real and imaginary parts of \(\Log\) are continuous, we conclude that \(\Log\) is continuous.
Fix a complex number \(w\text{.}\) Having established that \(\Log\) is continuous on \(\C-(-\infty, 0]\) it follows that the principal branch of \(z^w\) defined as
\begin{equation*} f(z)=e^{w\Log z} \end{equation*}
is continuous on its domain \(\C-(-\infty, 0]\text{.}\)
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