There are several methods for proving continuity:
- Concatenation Theorems: If the function can be written as a concatenation of continuous functions, it's continuous by the Concatenation Theorems.
- Using the local nature of continuity: If a function looks like another well-known continuous function in a small neighborhood of a point, then it must also be continuous in this point.
- Considering the left- and right-sided limits: If one can show that the left- and right-sided limits of a function are the same in some point, then the function is continuous in this point.
- Showing the sequence criterion: Using the sequence criterion means that the limit can be pulled into the function, i.e. we can consider the limit of the arguments. For a sequence
of arguments with limit value
it must hold
.
- Showing the epsilon-delta criterion: For every
it must be show that there exists some
such that for all arguments
with a distance smaller than
from the point
, the inequality
is satisfied.
Composition of continuous functions[Bearbeiten]
→ Main article: Composition of continuous functions
General proof sketch[Bearbeiten]
Following the concatenation theorems, every composition of continuous functions is again continuous function. So if
can be written as a concatenation of continuous functions, we can directly infer continuity of
. A corresponding proof could be of the following form:
In a lecture, this proof scheme may of course only be applied if the concatenation theorems have already been treated before. In any way, it is a very efficient method to characterize continuous functions.
Exercise (Epsilon-Delta proof for the continuity of the Square Root Function)
Show, using the epsilon-delta criterion, that the following function is continuous:
How to get to the proof? (Epsilon-Delta proof for the continuity of the Square Root Function)
We need to show, that for any given
, there is a
, such that all
with
satisfy the inequality
. So let us take a look at the target inequality
and estimate the absolute
from above. We are able to control the term
. Therefor, would like to get an upper bound for
including the expression
. So we are looking for an inequality of the form
Here,
is some expression depending on
and
. The second factor is smaller than
and can be made arbitrarily small by a suitable choice of
. Such a bound is constructed as follows:
Since
, there is:
If we now choose
small enough, such that
, then we obtain our target inequality
. But
still depends on
, so
would have to depend on , too - and we required one choice of
which is suitable for all
. THerefore, we need to get rid of the
-dependence. This is done by an estimate of the first factor, such that our inequality takes the form
:
We even made
independent of
, which would in fact not have been necessary. So we obtain the following inequality
We need the estimate
, in order to fulfill the target inequality
. The choice of
is sufficient for that. So let us write down the proof:
Using the Local Nature of Continuity[Bearbeiten]
If applicable, one can use the fact that continuity is a local property. Namely, when a function
looks the same as another function
in a certain point
, i.e.
on some small neighborhood of
, and we know
to be continuous, then
must also be continuous at
. For example, consider the function
with
for positive
and
for negative
. Now we choose some random positive number
. In a sufficiently small neighborhood of
,
is constant
:
Since constant functions are continuous,
is also continuous at the point
. Similarly one can show that
is also locally constant for negative number. Then
is also continuous for negative numbers, and is therefore continuous for all real numbers. In the proof we write:
Such an argument can often be applied to functions that are defined by a case differential. Our function
is a good example of that. In conclusion, it's defined as:
However, the local nature of continuity can not be used as an argument for every function defined by a case differential! Let's consider the following function
:
For all points that are not zero we can construct a proof like we've done earlier in this chapter to show that the function is continuous at these points. However, at the point
we have to construct a different argument. For example, here we could consider the left- and right-sided limits.
→ Main article: Sequential definition of continuity
Review: Sequence Criterion[Bearbeiten]
General Proof Structure[Bearbeiten]
In order to prove continuity of a function at some
, we need to show that for each sequence
of arguments converging to
, there is
. A proof for this could schematically look as follows:
In order to prove continuity for the function
(for all arguments in its domain of definition), we need to slightly adjust that scheme:
Exercise (Continuity of quadratic functions)
Show that the quadratic function
is continuous.
Proof (Continuity of quadratic functions)
Let
. Consider any sequence
, converging to
. There is
So we may pull the limit inside the brackets for the quadratic function and hence, it is continuous.
Epsilon-Delta Criterion [Bearbeiten]
→ Main article: Epsilon-delta definition of continuity
Review: Epsilon-Delta Criterion[Bearbeiten]
General Proof Structure[Bearbeiten]
In mathematical quantifiers, the epsilon-delta definition of continuity of a function
at the point
reads:
This technical method of writing the claim specifies the general proof structure for proving continuity using the epsilon-delta criterion:
Example Problem and General Procedure[Bearbeiten]
Exercise (Continuity of the quadratic function)
Prove that the function
with
is continuous.
How to get to the proof? (Continuity of the quadratic function)
For this proof, we need to show that the square function is continuous at any argument
. Using the proof structure for the epsilon-delta criterion, we are given an arbitrary
. Our job is to find a suitable
, such that the inequality
holds for all
.
In order to find a suitable
, we plug in the definition of the function
into the expression
which shall be smaller than
:
The expression
may easily be controlled by
. Hence, it makes sense to construct an upper estimate for
which includes
and a constant. The factor
appears if we perform a factorization using the third binomial formula:
The requirement
allows for an upper estimate of our expression:
The
we are looking for may only depend on
and
. So the dependence on
in the factor
is still a problem. We resolve it by making a further upper estimate for the factor
. We will use a simple, but widely applied "trick" for that: A
is subtracted and then added again at another place (so we are effectively adding a 0) , such that the expression
appears:
The absolute
is obtained using the triangle inequality. This absolute
is again bounded from above by
:
So reshaping expressions and applying estimates, we obtain:
With this inequality in hand, we are almost done. If
is chosen in a way that
, we will get the final inequality
. This
is practically found solving the quadratic equation
for
. Or even simpler, we may estimate
from above. We use that we may freely impose any condition on
. If we, for instance, set
, then
which simplifies things:
So
will also do the job. This inequality can be solved for
to get the second condition on
(the first one was
):
So any
fulfilling both conditions does the job:
and
have to hold. Ind indeed, both are true for
. This choice will be included into the final proof:
Concatenated absolute function[Bearbeiten]
Exercise (Example for a proof of continuity)
Prove that the following function is continuous at
:
How to get to the proof? (Example for a proof of continuity)
We need to show that for each given
, there is a
, such that for all
with
the inequality
holds. In our case,
. So by choosing
for
small enough, we may control the expression
. First, let us plug
into
in order to simplify the inequality to be shown
:
The objective is to "produce" as many expressions
as possible, since we can control
. It requires some experience with epsilon-delta proofs in order to "directly see" how this is achieved. First, we need to get rid of the double absolute. This is done using the inequality
. For instance, we could use the following estimate:
However, this is a bad estimate as the expression
no longer tends to 0 as
. To resolve this problem, we use
before applying the inequality
:
A factor of
can be directly extracted out of this with the third binomial fomula:
And we can control it by
:
Now, the required
must only depend on
and
. Therefore, we have to get rid of the
-dependence of
. This can be done by finding an upper bound for
which does not depend on
. As we are free to chose any
for our proof, we may also impose any condition to it which helps us with the upper bound. In this case,
turns out to be quite useful. In fact,
or an even higher bound would do this job, as well. What follows from this choice?
As before, there has to be
. As
, we now have
and as
, we obtain
and
. This is the upper bound we were looking for:
As we would like to show
, we set
. And get that our final inequality holds:
So if the two conditions for
are satisfied, we get the final inequality. In fact, both conditions will be satisfied if
, concluding the proof. So let's conclude our ideas and write them down in a proof:
Exercise (Continuity of the hyperbolic function)
Prove that the function
with
is continuous.
How to get to the proof? (Continuity of the hyperbolic function)
The basic pattern for epsilon-delta proofs is applied here, as well. We would like to show the implication
. Forst, let us plug in what we know and reshape our terms a bit until a
appears:
By assumption, there will be
, of which we can make use:
The choice of
may again only depend on
and
so we need a smart estimate for
in order to get rid of the
-dependence. To do so, we consider
.
Why was
and not
chosen? The explanation is quite simple: We need a
-neighborhood inside the domain of definition of
. If we had simply chosen
, we might have get kicked out of this domain. For instance, in case
, the following problem appears:
The biggest
-value with
is
and the smallest one is
. However,
is not inside the domain of definition as
. In particular,
is element of that interval, where
cannot be defined at all.
A smarter choice for
, such that the
-neighborhood doesn't touch the
-axis is half of the distance to it, i.e.
. A third of this distance or other fractions smaller than 1 would also be possible:
,
or
.
As we chose
and by
, there is
. This allows for an upper bound:
and we may write:
So we get the estimate:
Now, we want to prove
. Hence we choose
. Plugging this in, our final inequality
will be fulfilled.
So again, we imposed two conditions for
:
and
. Both are fulfilled by
, which we will use in the proof:
Concatenated square root function[Bearbeiten]
Exercise (Epsilon-Delta proof for the continuity of the Square Root Function)
Show, using the epsilon-delta criterion, that the following function is continuous:
How to get to the proof? (Epsilon-Delta proof for the continuity of the Square Root Function)
We need to show, that for any given
, there is a
, such that all
with
satisfy the inequality
. So let us take a look at the target inequality
and estimate the absolute
from above. We are able to control the term
. Therefor, would like to get an upper bound for
including the expression
. So we are looking for an inequality of the form
Here,
is some expression depending on
and
. The second factor is smaller than
and can be made arbitrarily small by a suitable choice of
. Such a bound is constructed as follows:
Since
, there is:
If we now choose
small enough, such that
, then we obtain our target inequality
. But
still depends on
, so
would have to depend on , too - and we required one choice of
which is suitable for all
. THerefore, we need to get rid of the
-dependence. This is done by an estimate of the first factor, such that our inequality takes the form
:
We even made
independent of
, which would in fact not have been necessary. So we obtain the following inequality
We need the estimate
, in order to fulfill the target inequality
. The choice of
is sufficient for that. So let us write down the proof:
Discontinuity of the topological sine function[Bearbeiten]
Exercise (Discontinuity of the topological sine fucntion)
Prove the discontinuity at
for the topological sine function:
How to get to the proof? (Discontinuity of the topological sine fucntion)
In this exercise, discontinuity has to be shown for a given function. This is done by the negation of the epsilon-delta criterion. Our objective is to find both an
and an
, such that
and
. Here,
may be chosen depending on
, while
has to be the same for all
. For a solution, we may proceed as follows
Step 1: Simplify the target inequality
Step 2: Choose a suitable 
We consider the graph of the function
. It will help us finding the building bricks for our proof:
We need to find an
, such that there are arguments in each arbitrarily narrow interval
whose function values have a distance larger than
from
. Visually, no matter how small the width
of the
-
-rectangle is chosen, there will always be some points below or above it.
Taking a look at the graph, we see that our function oscillates between
and
. Hence,
may be useful. In that case, there will be function values with
in every arbitrarily small neighborhood around
. We choose
. This is visualized in the following figure:
After
has been chosen, an arbitrary
will be assumed, for which we need to find a suitable
. This is what we will do next.
Step 3: Choice of a suitable 
Sequence Criterion: Absolute Value Function[Bearbeiten]
Exercise (Continuity of the absolute function)
Prove continuity for the absolute function.
Discontinuity of the topological sine function[Bearbeiten]
Exercise (Discontinuity of the topological sine function)
Prove discontinuity of the following function:
How to get to the proof? (Discontinuity of the topological sine function)
Discontinuity of
means that this function has at least one argument where
is discontinuous. For each
,
is equal to the functionn
in a sufficiently small neighbourhood of
. Since
is just a composition of continuous functions, it is continuous itself and therefore,
must be continuous for all
, as well. So we know that the discontinuity may only be situated at
.
In order to prove that
is discontinuous at
, we need to fincd a sequence of arguments
converging to
but with
. To find such a function, let us take a look at the graph of the function
:
In this figure, we recognize that
takes any value between
and
infinitely often in the vicinity of
. So, for instance, we may just choose
such that
is always
. This guarantees that
- and actually any other real number
between
and
in place of
would do the job. But we need to make a specific choice for
, and
is a very simple one. In addition, we will choose
to converges to zero from above.
The following figure also contains the sequence elements
beside our function
. We may clearly see that for
the sequence of function values converges to
, which is different from the function value
:
But what are the exact values of these
for which we would like to have
To answer this question, let us resolve the equation
for
:
So for each
with
, we have
. In order to get positive
converging to zero from above, we may for instance choose
. In that case:
And we have seen that
. So we found just a sequence of arguments
, which proves discontinuity of
at
.
Epsilon-Delta Criterion: Linear Function[Bearbeiten]
Exercise (Continuity of a linear function)
Prove that a linear function
with
is continuous.
Epsilon-Delta Criterion: Concatenated Absolute Value Function[Bearbeiten]
Exercise (Example for a proof of continuity)
Prove that the following function is continuous at
:
How to get to the proof? (Example for a proof of continuity)
We need to show that for each given
, there is a
, such that for all
with
the inequality
holds. In our case,
. So by choosing
for
small enough, we may control the expression
. First, let us plug
into
in order to simplify the inequality to be shown
:
The objective is to "produce" as many expressions
as possible, since we can control
. It requires some experience with epsilon-delta proofs in order to "directly see" how this is achieved. First, we need to get rid of the double absolute. This is done using the inequality
. For instance, we could use the following estimate:
However, this is a bad estimate as the expression
no longer tends to 0 as
. To resolve this problem, we use
before applying the inequality
:
A factor of
can be directly extracted out of this with the third binomial fomula:
And we can control it by
:
Now, the required