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8.20
Definition (Limit point.)
Let
![$S$](img427.gif)
be a subset of
![$\mbox{{\bf C}}$](img2.gif)
and let
![$a\in\mbox{{\bf C}}$](img86.gif)
. We say
![$a$](img590.gif)
is a
limit point of
![$S$](img427.gif)
if there is a sequence
![$f$](img13.gif)
in
![$S\backslash\{a\}$](img702.gif)
such that
![$f\to a$](img703.gif)
.
8.22
Exercise.
Supply the proof for Case 2 of example
8.21; i.e., show that
![$0$](img34.gif)
is
a limit point of
![$D(0,1)$](img705.gif)
.
8.23
Example.
The set
![$\mbox{{\bf Z}}$](img722.gif)
has no limit points. Suppose
![$\alpha\in\mbox{{\bf C}}$](img10.gif)
, and there is a sequence
![$f$](img13.gif)
in
![$\mbox{{\bf Z}}\backslash\{\alpha\}$](img723.gif)
such that
![$f\to\alpha$](img724.gif)
. Let
![$g(n)=f(n)-f(n+1)$](img725.gif)
for
all
![$n\in\mbox{{\bf N}}$](img9.gif)
. By the translation thoerem
![$g\to\alpha-\alpha=0$](img726.gif)
; i.e.,
![$g$](img111.gif)
is a
null sequence. Let
![$N_g$](img102.gif)
be a precision function for
![$g$](img111.gif)
. Then for all
![$n\in\mbox{{\bf N}}$](img9.gif)
,
Now
![$\vert f(n)-f(n+1)\vert\in\mbox{{\bf N}}$](img728.gif)
, so it follows that
and hence
This
contradicts the fact that
![$f(n)\in\mbox{{\bf Z}}\backslash\{\alpha\}$](img731.gif)
for all
![$n\in\mbox{{\bf N}}$](img9.gif)
.
8.24
Definition (Limit of a function.)
Let
![$f$](img13.gif)
be a complex function, and let
![$a$](img590.gif)
be a limit point of
![$\mbox{{\rm dom}}(f)$](img623.gif)
. We say
that
has a limit at ![$a$](img590.gif)
or that
exists if there exists a function
![$F$](img733.gif)
with
![$\mbox{{\rm dom}}(F)=\mbox{{\rm dom}}(f)\cup\{a\}$](img734.gif)
such that
![$F(z)=f(z)$](img735.gif)
for all
![$z\in\mbox{{\rm dom}}
(f)\backslash\{a\}$](img736.gif)
, and
![$F$](img733.gif)
is continuous at
![$a$](img590.gif)
. In this case we denote the value of
![$F(a)$](img737.gif)
by
![$\displaystyle {\lim_a f}$](img738.gif)
or
![$\displaystyle {\lim_{z\to a}f(z)}$](img739.gif)
.
Theorem
8.30 shows that this definition makes sense. We will
give some examples before proving that theorem.
8.25
Warning.
Notice that
![$\displaystyle {\lim_{a}f}$](img732.gif)
is defined only
when
![$a$](img590.gif)
is a limit point of
![$\mbox{{\rm dom}}(f)$](img623.gif)
. For each complex number
![$\beta$](img740.gif)
, define a function
![$F_\beta: \mbox{{\bf N}}\cup \{ {1\over 2}\} \to\mbox{{\bf C}}$](img741.gif)
by
Then
![$F_\beta$](img743.gif)
is continuous, and
![$F(n) = n!$](img744.gif)
for all
![$n\in\mbox{{\bf N}}$](img9.gif)
.
If I did not put the requirement that
![$a$](img590.gif)
be a limit point of
![$\mbox{{\rm dom}}(f)$](img623.gif)
in the above definition, I'd have
I certainly do not want this to be the case.
8.26
Example.
Let
![$\displaystyle {f(z)={{z^2-1}\over {z-1}}}$](img746.gif)
for all
![$z\in\mbox{{\bf C}}\backslash\{1\}$](img747.gif)
and let
![$F(z)=z+1$](img748.gif)
for all
![$z\in\mbox{{\bf C}}$](img66.gif)
. Then
![$f(z)=F(z)$](img749.gif)
on
![$\mbox{{\bf C}}\backslash\{1\}$](img750.gif)
and
![$F$](img733.gif)
is
continuous at
![$1$](img522.gif)
. Hence
![$\displaystyle {\lim_1 f=F(1)=2}$](img751.gif)
.
8.27
Example.
If
![$\displaystyle {f(z)=\cases{ z &for $\neq 1$\cr 3 &for $z=1$\cr}}$](img752.gif)
, then
![$\displaystyle {\lim_1f=1}$](img753.gif)
, since the function
![$F: z\mapsto z$](img754.gif)
agrees with
![$f$](img13.gif)
on
![$\mbox{{\bf C}}\backslash\{1\}$](img750.gif)
and is continuous at
![$1$](img522.gif)
.
8.28
Example.
If
![$f$](img13.gif)
is continuous at
![$a$](img590.gif)
, and
![$a$](img590.gif)
is a limit point of domain
![$f$](img13.gif)
,
then
![$f$](img13.gif)
has a limit at
![$a$](img590.gif)
, and
8.29
Example.
Let
![$\displaystyle {f(z)={{z^*}\over z}}$](img756.gif)
for all
![$z\in\mbox{{\bf C}}\setminus \{0\}$](img757.gif)
. Then
![$f$](img13.gif)
has no limit at
![$0$](img34.gif)
.
Proof: Suppose there were a continuous function
on
such that
on
. Let
and
. Then
and
and so
and also
Hence we get the contradiction
.
8.30
Theorem (Uniqueness of limits.)
Let
be a complex function, and let
be a limit point of
. Suppose
are two functions each having domain
, and each continuous
at
, and satisfying
for all
. Then
.
Proof:
is continuous at
, and
on
. Let
be a sequence in
such that
.
Since
is continuous at
, we have
i.e.,
so
; i.e.,
.
Proof: Suppose
has a limit at
, and let
be a continuous function with
, and
for all
.
Let
be a sequence in
such that
. Then
is a sequence in
, so by continuity of
,
Hence, condition (8.33) holds with
.
Conversely, suppose there is a number
such that
![\begin{displaymath}
\mbox{for every sequence $y$\ in
$\mbox{{\rm dom}}(f)\backsl...
...box{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f\circ y\to L.).
\end{displaymath}](img792.gif) |
(8.34) |
Define
by
I need to show that
is continuous at
. Let
be a sequence in
such that
. I want to show that
.
Let
be a sequence in
such that
. (Such a sequence exists because
is a limit point of
). Define a sequence
in
by
Let
and
be precision functions
for
and
respectively. Let
Then
is a precision function for
, since
for all
and all
,
Hence
, and by assumption (8.34), it follows
that
. I now claim that
, and in fact any
precision function
for
is a precision
function for
.
For all
and all
,
This completes the proof.
8.35
Example.
Let
I want to determine whether
![$f$](img13.gif)
has a limit at
![$0$](img34.gif)
, i.e., I want to know whether there is a number
![$L$](img28.gif)
such that for every sequence
![$z$](img795.gif)
in
If
![$x\in \mbox{{\bf R}}^+$](img816.gif)
and
![$\gamma \in \mbox{{\bf Q}}^+$](img817.gif)
then
Since
![$\vert 2-\gamma\vert$](img819.gif)
is either
![$2-\gamma$](img820.gif)
or
![$\gamma - 2$](img821.gif)
, we have
For each
![$\gamma \in \mbox{{\bf Q}}^+$](img817.gif)
, define a sequence
![$z_\gamma$](img823.gif)
by
Then
![$z_\gamma \to 0$](img825.gif)
, and
Hence
It follows that
![$f$](img13.gif)
has no limit at
![$0$](img34.gif)
.
Let
. It is clear that
maps points on the horizontal line
to
other points on the line
. I'll now look at the image of the parabola
under
.
So
![$f$](img13.gif)
maps the right half of the parabola
![$y=cx^2$](img830.gif)
into the vertical line
![$\displaystyle {x={c\over {1+c^2}}}$](img832.gif)
, and
![$f$](img13.gif)
maps the left half of the parabola to the line
![$\displaystyle {x={{-c}\over {1+c^2}}}$](img833.gif)
. Parabolas with
![$c>0$](img834.gif)
get mapped to the upper half
plane, and parabolas with
![$c<0$](img835.gif)
get mapped to the lower half plane. The
figure below
shows some parabolas and horizontal lines and their images under
![$f$](img13.gif)
.
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
![$y=cx^2$](img830.gif)
where
![$\vert c\vert \geq 1$](img837.gif)
. What do the images of the parabolas
![$y=cx^2$](img830.gif)
look
like when
![$\vert c\vert < 1$](img838.gif)
?)
8.37
Example.
It isn't quite true that `` the limit of the sum is the sum of the limits."
Let
Then from the continuity of the square root function and the composition theorem,
But
![$\displaystyle {\lim_0(f+g)}$](img841.gif)
does not exist, since
![$\mbox{{\rm dom}}(f+g)=\{0\}$](img842.gif)
and
![$0$](img34.gif)
is not a limit
point of
![$\mbox{{\rm dom}}(f+g)$](img843.gif)
.
Proof: Suppose that
and
exist. Let
be any
sequence in
such that
. Then
is a sequence
in both
and
, so
By the sum theorem for limits of sequences,
Hence
has a limit at
, and
.
The other parts of the theorem are proved similarly, and the proofs are left to
you.
8.39
Exercise.
Prove the product theorem for limits; i.e., show that if
![$f,g$](img109.gif)
are complex
functions such that
![$f$](img13.gif)
and
![$g$](img111.gif)
have limits at
![$a\in\mbox{{\bf C}}$](img86.gif)
, and if
![$a$](img590.gif)
is a limit point
of
![$\mbox{{\rm dom}}(f)\cap\mbox{{\rm dom}}(g)$](img844.gif)
, then
![$f\cdot g$](img653.gif)
has a limit at
![$a$](img590.gif)
and
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
for all
. Then
is
bounded on
since
However,
![$f$](img13.gif)
is not a bounded function, since
for all
![$n\in\mbox{{\bf Z}}_{\geq 1}$](img117.gif)
.
Let
(
![$F$](img733.gif)
is the real part of the discontinuous function from example
8.35.)
I claim
is bounded by
. For all
,
(NOTE:
![$\max(\vert a\vert,\vert b\vert)^2$](img874.gif)
is either
![$a^2$](img875.gif)
or
![$b^2$](img876.gif)
.) Hence if
![$(a,b)\neq(0,0)$](img877.gif)
, then
To prove my claim, apply this result with
![$a=x\vert x\vert$](img879.gif)
and
![$b=y$](img880.gif)
.
8.42
Exercise.
Show that
for all
![$(a,b)\in\mbox{{\bf R}}\times\mbox{{\bf R}}\backslash\{(0,0)\}$](img882.gif)
, and that equality holds if and only
if
![$\vert a\vert=\vert b\vert$](img883.gif)
. (This shows that
![$\displaystyle {{1\over 2}}$](img884.gif)
is a bound for the function
![$F$](img733.gif)
in
the previous example.) HINT: Consider
![$(\vert a\vert-\vert b\vert)^2$](img885.gif)
.
Next: 9. Properties of Continuous
Up: 8. Continuity
Previous: 8.2 Continuity
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