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# Find a function continuous at a single point of an interval

Find a function continuous at a single point of an interval, but discontinuous at all other points of the interval.

Consider the interval for any , and define

for all .

We claim is continuous at 0, but discontinuous at every other point of the interval.

Proof. First, we show is continuous at 0. To do this we must show that for all there exists a such that whenever . So, let be given. Then, choose . Then we have,

since from the definition of . Thus, is continuous at 0.

Now, we must show is discontinuous at every other point of the interval. Let be any nonzero point of the interval.

We consider two cases:
Case #1: is rational. Then . Since by assumption, we consider . Since the irrationals are dense in the reals (see Exercise #9 of Section I.3.12 here) we know that for any there is some irrational number such that . But then, . This means is not continuous at since we have found an such that there is no we can choose such that whenever .

Case #2: is irrational. Since is irrational we know . Consider . Since the rationals are dense in the reals (see Exercise #6 of Section I.3.12 here) we know that for any there exists some rational number such that . But then, . This means is not continuous at . (Again, because we have shown that there exists an such that for every there is some number such that with

(Note: There are much nicer ways to do this using sequences, but Apostol doesn’t develop those techniques, so we have to work straight from the epsilon-delta definitions.)

# Consider the continuity of x(-1)^[1/x]

Define:

where denotes the greatest integer function, or floor function. Sketch the graph of for and . Evaluate

Is it possible to define in a way that makes continuous at 0.

First, we sketch the graph of on the requested intervals.

As , .
As , .

If we define then is continuous at 0.

# Consider the continuity of (-1)^[1/x]

Define:

where denotes the greatest integer function, or floor function. Sketch the graph of for and . Evaluate

Is it possible to define in a way that makes continuous at 0.

First, we sketch the graph of on the requested intervals.

As , alternates between and .
As , alternates between and .

There is no way to define to make continuous at 0 since will take both values and no matter how small we choose our . (So, if we were to try to define , then for there is no such that whenever , and similarly if we try to define .)

# Consider the limit and continuity of the floor function of (1/x)

Define:

where denotes the greatest integer function, or floor function. Sketch the graph of for and . Evaluate

Is it possible to define in a way that makes continuous at 0.

First, we sketch the graph of on the requested intervals.

As , takes on arbitrarily large positive values.
As , takes on arbitrarily large negative values.

There is no way to define to make continuous at 0.

# Prove that sin (1/x) has no limit as x approaches 0

Define the function:

Prove there is no such that

Proof. We prove this by contradiction. Suppose there does exist some such that . From the definition of limit this means that for all there exists a such that

First, we claim that for any such we must have . This must be the case since if then , so we may choose such that . But then, since for all , we have

(Where we used part (i) this exercise (Section I.4.9, Exercise #1) for the inequality .) This contradicts our choice of , so must be less than or equal to 1.
Next, suppose . Then choose . To obtain our contradiction we must show that there is no such that

By the Archimedean property of the real numbers we know that for any , there exists a positive integer with such that . (The Archimedean property guarantees us an integer, but then the inequality will hold for any larger integer, so we can choose one with the property of being .) But,

Then, from the definition of and since and we have,

But then,

Furthermore,

This contradicts that if then for every we have for all . In other words, we found an greater than 0 (in particular, we found ) such that no matter how small we choose there exists an such that is smaller than , but is bigger than . This exactly contradicts the definition of . Hence, there can be no such number

This proof shows that there is now way to define so that is continuous at 0. We see this since for to be continuous at 0 we must have . But if we try to define for any , we will fail to achieve continuity since the proof shows for any we might want to choose.

# Give an alternate proof of the continuity of the sine and cosine functions

1. Use the inequality

to prove that the sine function is continuous at 0.

2. Recall the trig identity,

Use this and part (a) to prove that the cosine function is continuous at 0.

3. Use the formulas for sine and cosine of a sum to prove that the sine and cosine functions are continuous for all .

1. Proof. To show is continuous at 0, we must show that . We show this limit is zero directly from the epsilon-delta definition of the limit, i.e., given arbitrary positive , we have whenever . Let be an arbitrary number greater than 0, . Then, let . Using the given inequality we have,

Thus, is continuous at

2. Proof. First, using the given trig identity we have

Thus,

Thus, cosine is continuous at 0.

3. Proof. Finally, to show sine and cosine are continuous for all , we show that , and . First, we recall the formulas for the sine and cosine of a sum,

So, we compute the limits

Therefore, sine and cosine are continuous for all

# Consider the continuity of (tan x)/x at the point x=0

Consider the function

Determine what happens to as . Define so that will be continuous at , if possible.

First, we sketch the graph of on the intervals and .

To determine what happens to as we can take the limit, (recalling that ),

Since is not defined at it is not continuous there. However, since the limit exists we could redefine by

Then, this new (which has the same values as the original function everywhere the original function is defined, but has the additional feature of being defined at ) is continuous at since it is defined there and .

Note what we have done here. We had a function which was undefined at the point . We created a new function (which we also called , which is a bit confusing) that took the same values as the original function for all , and took the value 1 when . In this way we “removed” the discontinuity of the original at the point , but we should be aware that this is actually a new function (since it has a different domain than the original).

# Find all points at which tangent and cotangent are continuous

Find the points at which is continuous and at which is continuous.

Since

and , are continuous everywhere we know (by Theorem 3.2) that is continuous everywhere is not zero. We proved in this exercise (Section 2.8, #1 (b) of Apostol) that

Thus, is continuous for

Similarly,

and by Theorem 3.2 we then have is continuous everywhere that is not zero. We proved in the same exercise (Section 2.8, #1(a) of Apostol) that

Hence, is continuous for

# Find a constant such that a given piecewise function is continuous everywhere

Given constants define:

If are fixed find all values of such that is continuous at .

By the definition of continuity, we know that continuous at means that is defined and . Since and are defined for all , we know that is defined for every . Then, to show that it is continuous we must show

From the definition of we know

Then, taking the limit as approaches from the right (since since is continuous and as approaches from the left),

So, we must have

If then,

# Find a constant such that a given piecewise function is continuous

For a constants , define:

If are fixed, find all values for such that is continuous at .

By the definition of continuity of a function at a point, we know that is continuous at means is defined at , and .

Since and are defined for all , we know that is defined at for all . So, we must then find values of such that

From the definition of , we know

Then, we evaluate the limit as through values greater than (since the limit as through values less than is since is a continuous function, and for values less than , ),

Thus, for to be continuous at we must have,

If , then we have