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# 8.3 Limits

8.20   Definition (Limit point.) Let be a subset of and let . We say is a limit point of if there is a sequence in such that .

8.21   Example. Let be the unit disc, and let . We'll show that is a limit point of if and only if .

Proof that ( is a limit point of .

Suppose is a limit point of . Then there is a sequence in such that . Since the absolute value function is continuous, it follows that . Since we know that (and hence .) for all . By the inequality theorem for limits of sequences, , i.e. .

Proof that is a limit point of .

Case 1:
Suppose . Let for all . Then so , and clearly . Now , so is a limit point of .
Case 2:
. This case is left to you.

8.22   Exercise. Supply the proof for Case 2 of example 8.21; i.e., show that is a limit point of .

8.23   Example. The set has no limit points. Suppose , and there is a sequence in such that . Let for all . By the translation thoerem ; i.e., is a null sequence. Let be a precision function for . Then for all ,

Now , so it follows that

and hence

This contradicts the fact that for all .

8.24   Definition (Limit of a function.) Let be a complex function, and let be a limit point of . We say that has a limit at or that exists if there exists a function with such that for all , and is continuous at . In this case we denote the value of by or . Theorem 8.30 shows that this definition makes sense. We will give some examples before proving that theorem.

8.25   Warning. Notice that is defined only when is a limit point of . For each complex number , define a function by

Then is continuous, and for all . If I did not put the requirement that be a limit point of in the above definition, I'd have

I certainly do not want this to be the case.

8.26   Example. Let for all and let for all . Then on and is continuous at . Hence .

8.27   Example. If , then , since the function agrees with on and is continuous at .

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

8.29   Example. Let for all . Then has no limit at .

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., .

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 that the implied limit exists.
a)
b)
c)
d)
(Here ).
e)

8.32   Theorem. Let be a complex function and let be a limit point of . Then has a limit at if and only if there exists a number in such that for every sequence in
 (8.33)

In this case, .

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

 (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 has a limit at , i.e., I want to know whether there is a number such that for every sequence in

If and then

Since is either or , we have

For each , define a sequence by

Then , and

Hence

It follows that has no limit at .

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 maps the right half of the parabola into the vertical line , and maps the left half of the parabola to the line . Parabolas with get mapped to the upper half plane, and parabolas with get mapped to the lower half plane. The figure below shows some parabolas and horizontal lines and their images under .

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 where . What do the images of the parabolas look like when ?)

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 does not exist, since and is not a limit point of .

8.38   Theorem (Sum and product theorem.) Let be complex functions and let be a limit point of . If and exist, then , and all exist and

If is a limit point of and then exists and .

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 are complex functions such that and have limits at , and if is a limit point of , then has a limit at and

8.40   Definition (Bounded set and function.) A subset of is bounded if is contained in some disc ; i.e., if there is a number in such that for all . We call such a number a bound for .

Now suppose is a functiom from some set to and is a subset of . We say is bounded on if is a bounded set, and any bound for is called a bound for on . Thus a number is a bound for on if and only if

We say is bounded if is bounded on . If is not bounded on , we say is unbounded on .

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, is not a bounded function, since

for all .

Let

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

I claim is bounded by . For all ,

(NOTE: is either or .) Hence if , then

To prove my claim, apply this result with and .

8.42   Exercise. Show that

for all , and that equality holds if and only if . (This shows that is a bound for the function in the previous example.) HINT: Consider .

8.43   Exercise. A For each of the functions below:
1) Decide whether is bounded, and if it is, find a bound for .
2) Decide whether is bounded on , and if it is, find a bound for on .
3) Decide whether has a limit at , and if it does, find .
Here .
a)
for all .
b)
for all .
c)
for all .
d)
for all .
e)
for all .

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