8.3 Limits

8.20   Definition (Limit point.) Let $S$ be a subset of $\mbox{{\bf C}}$ and let $a\in\mbox{{\bf C}}$. We say $a$ is a limit point of $S$ if there is a sequence $f$ in $S\backslash\{a\}$ such that $f\to a$.

8.21   Example. Let $D(0,1)=\{z\in\mbox{{\bf C}}\colon\vert z\vert<1\}$ be the unit disc, and let $\alpha\in\mbox{{\bf C}}$. We'll show that $\alpha$ is a limit point of $D(0,1)$ if and only if $\vert\alpha\vert\leq 1$.

Proof that ($\alpha$ is a limit point of $D(0,1)) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$} \vert\alpha\vert \leq 1$.

Suppose $\alpha$ is a limit point of $D(0,1)$. Then there is a sequence $\{a_n\}$ in $D(0,1) \setminus \{\alpha\}$ such that $\{a_n\} \to \alpha$. Since the absolute value function is continuous, it follows that $\{\vert a_n\vert\} \to \vert\alpha\vert$. Since $a_n \in D(0,1)$ we know that $\vert a_n\vert< 1$ (and hence $\vert a_n\vert \leq 1$.) for all $n\in\mbox{{\bf N}}$. By the inequality theorem for limits of sequences, $\lim\{\vert a_n\vert\} \leq 1$, i.e. $\vert\alpha\vert\leq 1$.

Proof that $(\vert\alpha\vert \leq 1) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\alpha$ is a limit point of $D(0,1)$.

Case 1:

Suppose $0<\vert\alpha\vert\leq 1$. Let $${f_\alpha (n)={n\over {n+1}}\alpha}$$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$. Then $${\vert f_\alpha(n)\vert={n\over {n+1}}\vert\alpha\vert\leq {n\over {n+1}}<1}$$ so $f_\alpha(n)\in D(0,1)$, and clearly $f_\alpha(n)\neq \alpha$. Now $${ \{f_\alpha(n)\}_{n\geq 1}=\left\{ {1\over {1+{1\over n}}}\cdot\alpha\right\}_{n\geq 1}\to\alpha}$$, so $\alpha$ is a limit point of $D(0,1)$.

Case 2:

$\alpha=0$. This case is left to you.

8.22   Exercise. Supply the proof for Case 2 of example 8.21; i.e., show that $0$ is a limit point of $D(0,1)$.

8.23   Example. The set $\mbox{{\bf Z}}$ has no limit points. Suppose $\alpha\in\mbox{{\bf C}}$, and there is a sequence $f$ in $\mbox{{\bf Z}}\backslash\{\alpha\}$ such that $f\to\alpha$. Let $g(n)=f(n)-f(n+1)$ for all $n\in\mbox{{\bf N}}$. By the translation thoerem $g\to\alpha-\alpha=0$; i.e., $g$ is a null sequence. Let $N_g$ be a precision function for $g$. Then for all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} n\geq N_g\left({1\over 2}\right) &\mbox{$\Longrightarrow$}& \v... ...\ &\mbox{$\Longrightarrow$}& \vert f(n)-f(n+1)\vert<{1\over 2}. \end{eqnarray*}$$

Now $\vert f(n)-f(n+1)\vert\in\mbox{{\bf N}}$, so it follows that

$$n\geq N_g\left({1\over 2}\right)\mbox{$\hspace{1ex}\Longright... ...rt=0\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f(n)=f(n+1)$$

and hence

$$\displaystyle {\alpha=\lim f=f\left(N_g\left({1\over 2}\right)\right)}.$$

This contradicts the fact that $f(n)\in\mbox{{\bf Z}}\backslash\{\alpha\}$ for all $n\in\mbox{{\bf N}}$. $\mid\!\mid\!\mid$

8.24   Definition (Limit of a function.) Let $f$ be a complex function, and let $a$ be a limit point of $\mbox{{\rm dom}}(f)$. We say that $f$ has a limit at $a$ or that $${\lim_{a}f}$$ exists if there exists a function $F$ with $\mbox{{\rm dom}}(F)=\mbox{{\rm dom}}(f)\cup\{a\}$ such that $F(z)=f(z)$ for all $z\in\mbox{{\rm dom}} (f)\backslash\{a\}$, and $F$ is continuous at $a$. In this case we denote the value of $F(a)$ by $${\lim_a f}$$ or $${\lim_{z\to a}f(z)}$$. Theorem 8.30 shows that this definition makes sense. We will give some examples before proving that theorem.

8.25   Warning. Notice that $${\lim_{a}f}$$ is defined only when $a$ is a limit point of $\mbox{{\rm dom}}(f)$. For each complex number $\beta$, define a function $F_\beta: \mbox{{\bf N}}\cup \{ {1\over 2}\} \to\mbox{{\bf C}}$ by

$$F_\beta(n) = \cases{n! & if $n\in\mbox{{\bf N}}$,\cr \beta& if $n = {1\over 2}$,}$$

Then $F_\beta$ is continuous, and $F(n) = n!$ for all $n\in\mbox{{\bf N}}$. If I did not put the requirement that $a$ be a limit point of $\mbox{{\rm dom}}(f)$ in the above definition, I'd have

$$\lim_{n\to {1\over 2}} n! = F_\beta({1\over 2}) =\beta \mbox{ for all }\beta \in\mbox{{\bf C}}.$$

I certainly do not want this to be the case.

8.26   Example. Let $${f(z)={{z^2-1}\over {z-1}}}$$ for all $z\in\mbox{{\bf C}}\backslash\{1\}$ and let $F(z)=z+1$ for all $z\in\mbox{{\bf C}}$. Then $f(z)=F(z)$ on $\mbox{{\bf C}}\backslash\{1\}$ and $F$ is continuous at $1$. Hence $${\lim_1 f=F(1)=2}$$.

8.27   Example. If $${f(z)=\cases{ z &for $\neq 1$\cr 3 &for $z=1$\cr}}$$, then $${\lim_1f=1}$$, since the function $F: z\mapsto z$ agrees with $f$ on $\mbox{{\bf C}}\backslash\{1\}$ and is continuous at $1$.

8.28   Example. If $f$ is continuous at $a$, and $a$ is a limit point of domain $f$, then $f$ has a limit at $a$, and

$$\lim_{a}f = f(a).$$

8.29   Example. Let $${f(z)={{z^*}\over z}}$$ for all $z\in\mbox{{\bf C}}\setminus \{0\}$. Then $f$ has no limit at $0$.

Proof: Suppose there were a continuous function $F$ on $\mbox{{\bf C}}$ such that $F(z)=f(z)$ on $\mbox{{\bf C}}\backslash\{0\}$. Let $${\{a_n\}=\left\{ {i\over {n+1}}\right\}}$$ and $${\{b_n\}=\left\{ {1\over {n+1}}\right\}}$$. Then $\{a_n\}\to 0$ and $\{b_n\}\to 0$ and so

$$F(0)=\lim \{F(a_n)\}=\lim\left\{ { {{-i}\over {n+1}}\over {i\over {n+1}} }\right\}=\lim\{-1\}=-1$$

and also

$$F(0)=\lim\{F(b_n)\}=\lim \left\{ { {{1}\over {n+1}}\over {1\over {n+1}} }\right\}=\lim\{1\}=1.$$

Hence we get the contradiction $-1=1$. $\mid\!\mid\!\mid$

8.30   Theorem (Uniqueness of limits.) Let $f$ be a complex function, and let $a$ be a limit point of $\mbox{{\rm dom}}(f)$. Suppose $F,G$ are two functions each having domain $\mbox{{\rm dom}}(f)\cup\{a\}$, and each continuous at $a$, and satisfying $f(z)$ $=F(z)$ $=G(z)$ for all $z\in\mbox{{\rm dom}} (f)\backslash\{a\}$. Then $F(a)=G(a)$.

Proof: $F-G$ is continuous at $a$, and $F-G=0$ on $\mbox{{\rm dom}}(f)\backslash\{a\}$. Let $\{a_n\}$ be a sequence in $\mbox{{\rm dom}}(f)\backslash\{a\}$ such that $\{a_n\}\to a$. Since $F-G$ is continuous at $a$, we have

$$\{(F-G)(a_n)\}\to (F-G)(a);$$

i.e.,

$$\{0\}\to F(a)-G(a),$$

so $F(a)-G(a)=0$; i.e., $F(a)=G(a)$. $\mid\!\mid\!\mid$

8.31   Exercise. Investigate the following limits. (Give detailed reasons for your answers). In this exercise you should not conclude from the fact that I've written $${\lim_{w\to b} f(w)}$$ that the implied limit exists.

a)

$${ \lim_{t\to 4} t^{1\over 2}}.$$

b)

$${ \lim_{n\to 2} n!}.$$

c)

$${ \lim_{z \to 0} \vert z\vert^2 \left({1\over z} - {1\over z^*}\right).}$$

d)

$${ \lim_{z \to a} { {1\over z} - {1\over a} \over z-a}.}$$ (Here $a \in \mbox{{\bf C}}\setminus \{0\}$).

e)

$${ \lim_{t\to 0} {\sqrt{t+4} - 2 \over t}.}$$

8.32   Theorem. Let $f$ be a complex function and let $a$ be a limit point of $\mbox{{\rm dom}}(f)$. Then $f$ has a limit at $a$ if and only if there exists a number $L$ in $\mbox{{\bf C}}$ such that for every sequence $y$ in $\mbox{{\rm dom}}(f)\backslash\{a\}$

$$y\to a\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f\circ y\to L.$$ (8.33)

In this case, $${L=\lim_a f}$$.

Proof: Suppose $f$ has a limit at $a$, and let $F$ be a continuous function with $\mbox{{\rm dom}}(F)=\mbox{{\rm dom}}(f)\cup\{a\}$, and $F(z)=f(z)$ for all $z\in\mbox{{\rm dom}} (f)\backslash\{a\}$. Let $y$ be a sequence in $\mbox{{\rm dom}}(f)\backslash\{a\}$ such that $\{y_n\}\to a$. Then $y$ is a sequence in $\mbox{{\rm dom}}(F)$, so by continuity of $F$,

$$\{f(y_n)\}=\{F(y_n)\}\to F(a).$$

Hence, condition (8.33) holds with $L=F(a)$.

Conversely, suppose there is a number $L$ such that

$$\mbox{for every sequence $y$\ in $\mbox{{\rm dom}}(f)\backsl... ...box{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f\circ y\to L.).$$ (8.34)

Define $F\colon\mbox{{\rm dom}}(f)\cup\{a\}\to\mbox{{\bf C}}$ by

$$F(z)=\cases{ f(z) &if $z\in\mbox{{\rm dom}}(f)\backslash\{a\}$\cr L &if $z=a$.\cr}$$

I need to show that $F$ is continuous at $a$. Let $z$ be a sequence in $\mbox{{\rm dom}}(F)$ such that $z\to a$. I want to show that $F\circ z\to L$.

Let $w$ be a sequence in $\mbox{{\rm dom}}(f) \setminus \{a\}$ such that $w \to a$. (Such a sequence exists because $a$ is a limit point of $\mbox{{\rm dom}}(f)$). Define a sequence $y$ in $\mbox{{\rm dom}}(f) \setminus \{a\}$ by

$$y(n) = \cases{ z(n) & if $z(n) \neq a$\cr w(n) & if $z(n) = a$.}$$

Let $N_{z - \tilde{a}}$ and $N_{w - \tilde{a}}$ be precision functions for $z - \tilde{a}$ and $w-\tilde{a}$ respectively. Let

$$M(\varepsilon) = \max (N_{z- \tilde{a}}(\varepsilon), N_{w- \... ... \mbox{ for all }\varepsilon \in \mbox{${\mbox{{\bf R}}}^{+}$}.$$

Then $M$ is a precision function for $y - \tilde{a}$, since for all $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$ and all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} \lefteqn{ n \geq M(\varepsilon)}\\ &\mbox{$\Longrightarrow$}&... ...hspace{1ex}$}\vert y(n) - a\vert < \varepsilon & if $z(n) = a$.} \end{eqnarray*}$$

Hence $y \to a$, and by assumption (8.34), it follows that $f \circ y \to L$. I now claim that $F\circ z\to L$, and in fact any precision function $P$ for $f \circ y - \tilde L$ is a precision function for $F \circ z - \tilde{L}$. For all $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$ and all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} \lefteqn{n \geq P(\varepsilon) \mbox{$\hspace{1ex}\Longrightar... ...L\vert = \vert F(a) - L\vert = 0 < \varepsilon & if $z(n) = a.$} \end{eqnarray*}$$

This completes the proof. $\mid\!\mid\!\mid$

8.35   Example. Let

$$f(z)=f(x+iy) = f((x,y))={{xy\vert x\vert}\over {x^4+y^2}}+iy\mbox{ for all } z\in\mbox{{\bf C}}\backslash\{0\}.$$

I want to determine whether $f$ has a limit at $0$, i.e., I want to know whether there is a number $L$ such that for every sequence $z$ in $\mbox{{\bf C}}\backslash\{0\}$

$$z\to 0\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f(z)\to L.$$

If $x\in \mbox{{\bf R}}^+$ and $\gamma \in \mbox{{\bf Q}}^+$ then

$$f((x,x^\gamma)) = {x\cdot x\cdot x^\gamma \over x^4+x^{2\gamm... ...= {x^{2-\gamma} \over x^{2(2-\gamma)} + 1 } +ix^\gamma } & }$$

Since $\vert 2-\gamma\vert$ is either $2-\gamma$ or $\gamma - 2$, we have

$$f((x,x^\gamma)) = {x^{\vert 2-\gamma\vert} \over 1+x^{2\vert 2-\gamma\vert}} + ix^\gamma.$$

For each $\gamma \in \mbox{{\bf Q}}^+$, define a sequence $z_\gamma$ by

$$z_\gamma:n \mapsto \left( {1\over n},{1\over n^\gamma}\right) \mbox{ for all }n \in\mbox{${\mbox{{\bf Z}}}^{+}$}.$$

Then $z_\gamma \to 0$, and

$$f(z_\gamma(n)) ={ {1\over n^{\vert 2-\gamma\vert}} \over 1+{1\over n^{2\vert 2-\gamma\vert}}} +{i \over n^\gamma}.$$

Hence

$$\begin{eqnarray*} f\circ z_{\gamma} \to 0 &\mbox{if}& \gamma \neq 2\\ f \circ z_{\gamma} \to {1\over 2} &\mbox{if}& \gamma = 2. \end{eqnarray*}$$

It follows that $f$ has no limit at $0$.

Let $y_0\in\mbox{{\bf R}}$. It is clear that $f$ maps points on the horizontal line $y=y_0$ to other points on the line $y=y_0$. I'll now look at the image of the parabola $y=cx^2$ under $f$.

$$f(x+icx^2)={{xcx^2\vert x\vert}\over {x^4+c^2x^4}}+icx^2={{\v... ...over x}\left( {c\over {1+c^2}}\right)+icx^2\mbox{ for }x\neq 0.$$

So $f$ maps the right half of the parabola $y=cx^2$ into the vertical line $${x={c\over {1+c^2}}}$$, and $f$ maps the left half of the parabola to the line $${x={{-c}\over {1+c^2}}}$$. Parabolas with $c>0$ get mapped to the upper half plane, and parabolas with $c<0$ get mapped to the lower half plane. The figure below shows some parabolas and horizontal lines and their images under $f$.

8.36   Entertainment. Explain how the cat's nose in the above picture gets stretched, while its cheeks get pinched to a point. (Hint: The figure shows the images of some parabolas $y=cx^2$ where $\vert c\vert \geq 1$. What do the images of the parabolas $y=cx^2$ look like when $\vert c\vert < 1$?)

8.37   Example. It isn't quite true that `` the limit of the sum is the sum of the limits." Let

$$\begin{eqnarray*} f(x)&=&\sqrt x \mbox{ for } x\in [0,\infty) \\ g(x)&=&\sqrt{-x} \mbox{ for } x\in (-\infty,0].\\ \end{eqnarray*}$$

Then from the continuity of the square root function and the composition theorem,

$$\lim_0 f = 0 =\lim_0 g.$$

But $${\lim_0(f+g)}$$ does not exist, since $\mbox{{\rm dom}}(f+g)=\{0\}$ and $0$ is not a limit point of $\mbox{{\rm dom}}(f+g)$.

8.38   Theorem (Sum and product theorem.) Let $f,g$ be complex functions and let $a$ be a limit point of $\mbox{{\rm dom}}(f)\cap\mbox{{\rm dom}}(g)$. If $${\lim_a f}$$ and $${\lim_a g}$$ exist, then $${\lim_a(f+g)}$$, $${\lim_a(f-g)}$$ and $${\lim_a(f\cdot g)}$$ all exist and

$$\begin{eqnarray*} \lim_a(f+g)&=&\lim_a f+\lim_a g,\\ \lim_a(f-g)&=&\lim_af-\lim_ag,\\ \lim_a(f\cdot g)&=&\lim_af\cdot\lim_a g.\\ \end{eqnarray*}$$

If $a$ is a limit point of $${\mbox{{\rm dom}}\left({f\over g}\right)}$$ and $${\lim_ag\neq 0}$$ then $${\lim_a{f\over g}}$$ exists and $${\lim_a\left({f\over g}\right)={\displaystyle {\lim_af}\over \displaystyle {\lim_ag}}}$$.

Proof: Suppose that $${\lim_a f}$$ and $${\lim_a g}$$ exist. Let $x$ be any sequence in
$\mbox{{\rm dom}}(f+g)\backslash\{a\}$ such that $x\to a$. Then $x$ is a sequence in both $\mbox{{\rm dom}}(f)$ and $\mbox{{\rm dom}}(g)$, so

$$\lim\{f(x_n)\}=\lim_af \mbox{ and }\lim\{g(x_n)\}=\lim_ag.$$

By the sum theorem for limits of sequences,

$$\lim\{(f+g)(x_n)\}=\lim\{f(x_n)\}+\lim\{g(x_n)\}=\lim_af+\lim_ag.$$

Hence $f+g$ has a limit at $a$, and $${\lim_a(f+g)=\lim_af+\lim_ag}$$.

The other parts of the theorem are proved similarly, and the proofs are left to you. $\mid\!\mid\!\mid$

8.39   Exercise. Prove the product theorem for limits; i.e., show that if $f,g$ are complex functions such that $f$ and $g$ have limits at $a\in\mbox{{\bf C}}$, and if $a$ is a limit point of $\mbox{{\rm dom}}(f)\cap\mbox{{\rm dom}}(g)$, then $f\cdot g$ has a limit at $a$ and

$$\lim_a(f\cdot g)=\lim_af \cdot \lim_ag.$$

8.40   Definition (Bounded set and function.) A subset $S$ of $\mbox{{\bf C}}$ is bounded if $S$ is contained in some disc $\bar D(0,B)$; i.e., if there is a number $B$ in $\mbox{${\mbox{{\bf R}}}^{+}$}$ such that $\vert s\vert\leq B$ for all $s\in S$. We call such a number $B$ a bound for $S$.

Now suppose $f\colon U\to\mbox{{\bf C}}$ is a functiom from some set $U$ to $\mbox{{\bf C}}$ and $A$ is a subset of $U$. We say $f$ is bounded on $A$ if $f(A)$ is a bounded set, and any bound for $f(A)$ is called a bound for $f$ on $A$. Thus a number $B\in\mbox{${\mbox{{\bf R}}}^{+}$}$ is a bound for $f$ on $A$ if and only if

$$\vert f(a)\vert\leq B \mbox{ for all } a\in A.$$

We say $f$ is bounded if $f$ is bounded on $\mbox{{\rm dom}}(f)$. If $f$ is not bounded on $A$, we say $f$ is unbounded on $A$.

8.41   Examples. The definition of bounded sequence given in 7.41 is a special case of the definition just given for bounded function.

Let $${f(z)={1\over {1+z^2}}}$$ for all $z\in\mbox{{\bf C}}\backslash\{\pm i\}$. Then $f$ is bounded on $\mbox{{\bf R}}$ since

$$\vert f(z)\vert\leq{1\over {1+z^2}}\leq 1\mbox{ for all } z\in\mbox{{\bf R}}.$$

However, $f$ is not a bounded function, since

$$\begin{eqnarray*} \left\vert f\left(i+{1\over n}\right)\right\vert&=&\left\vert ... ... &=&{n\over {\sqrt{4+{1\over {n^2}}}}}\geq {n\over {\sqrt 5}}\\ \end{eqnarray*}$$

for all $n\in\mbox{{\bf Z}}_{\geq 1}$.

Let

$$F(z)=\cases{ \displaystyle {{{xy\vert x\vert}\over {x^4+y^2}} }&for $z\in\mbox{{\bf C}}\backslash\{0\}$ \cr 0 &for $z=0$.\cr}$$

($F$ is the real part of the discontinuous function from example 8.35.)

I claim $F$ is bounded by $1$. For all $a,b\in\mbox{{\bf R}}$,

$$\vert a\vert\;\vert b\vert\leq\max(\vert a\vert,\vert b\vert)^2\leq a^2+b^2.$$

(NOTE: $\max(\vert a\vert,\vert b\vert)^2$ is either $a^2$ or $b^2$.) Hence if $(a,b)\neq(0,0)$, then

$$\left\vert{{ab}\over {a^2+b^2}}\right\vert\leq 1.$$

To prove my claim, apply this result with $a=x\vert x\vert$ and $b=y$.

8.42   Exercise. Show that

$$\left\vert {{ab}\over {a^2+b^2}}\right\vert\leq{1\over 2}$$

for all $(a,b)\in\mbox{{\bf R}}\times\mbox{{\bf R}}\backslash\{(0,0)\}$, and that equality holds if and only if $\vert a\vert=\vert b\vert$. (This shows that $${{1\over 2}}$$ is a bound for the function $F$ in the previous example.) HINT: Consider $(\vert a\vert-\vert b\vert)^2$.

8.43   Exercise. A For each of the functions $f$ below:

1) Decide whether $f$ is bounded, and if it is, find a bound for $f$.

2) Decide whether $f$ is bounded on $\mbox{{\rm dom}}(f)\cap D(0,1)$, and if it is, find a bound for $f$ on $\mbox{{\rm dom}}(f)\cap D(0,1)$.

3) Decide whether $f$ has a limit at $0$, and if it does, find $${\lim_0 f}$$.

Here $z=(x,y)=x+iy$.

a)

$${f(z)={{x^2}\over {x^2+y^2}}}$$ for all $z\in\mbox{{\bf C}}\backslash\{0\}$.

b)

$${f(z)={{x^2y}\over {x^2+y^2}}}$$ for all $z\in\mbox{{\bf C}}\backslash\{0\}$.

c)

$${f(z)={(z^*)^2\over {z^2}} }$$ for all $z\in\mbox{{\bf C}}\backslash\{0\}$.

d)

$${f(x)={{x^2+x^6}\over {x^2+x^4}}}$$ for all $x\in\mbox{{\bf R}}\backslash\{0\}$.

e)

$${f(x)={{\sqrt{x+1}-1}\over x}}$$ for all $x\in[-1,\infty)\backslash\{0\}$.