### Video Transcript

Continuity of Functions

In this lesson, weβll learn how to
check the continuity of a function and to determine the interval on which a function
is continuous.

Now, you may be familiar with
finding the continuity of a function at a point and, indeed, the different types of
discontinuity that we come across. We can test whether a function is
continuous at a point using the following formal definition. A function π of π₯ is continuous
at the point where π₯ is equal to π if the limit, as π₯ approaches π of π of π₯,
is equal to the function evaluated at π₯ equals π.

Now, the so-much-implied
requirement to this condition is that both of these things must exist. In order for the normal limit to
exist, the left and the right limit must also exist as π₯ approaches π and be equal
to some value πΏ. Alongside this, π of π must be
defined, and it must be equal to the left, the right, and the normal limit for
continuity. Here, weβve said this value is
πΏ.

Now, letβs think about continuity
over an interval. The colloquial definition of this
might be, if youβre able to draw the graph of the function over the interval without
lifting your pen. A more rigorous way to think about
this would be to say, the function π of π₯ is continuous over an interval if the
requirement for continuity at a point holds for all values of π₯ within the
interval. Now, this may seem obvious, but one
way to check for continuity over an interval is to ensure that there are no
discontinuities in said interval. We can look at a couple of example
graphs to get a visual understanding of this.

Determine whether the following
statement is true or false. The function represented by the
graph is a continuous function.

For this question, weβve been given
a function π of π₯, which is defined for all values of π₯ over the real numbers,
denoted by these arrows. Now, we can almost immediately see
that our function π of π₯ is discontinuous, since at a value of π₯ equals three, we
have a gap in our graph. In fact, we may recognise this as a
jump discontinuity, with π of π₯ being undefined at the point three, two, denoted
by the hollow dot, and defined at the point three, one, denoted by the filled
dot.

If we were to look at the left- and
right-sided limits, as π₯ approaches three, we would find that although both exist,
their values disagree. And this would mean that the normal
limit does not exist. Here, we recall our definition for
continuity at a point, which says that the limit, as π₯ approaches π, of said
function must be equal to the value of the function evaluated where π₯ is equal to
π. Now, in our case, the limit, as π₯
approaches three of π of π₯, is not equal to π of three, since the limit does not
exist.

We have, therefore, proved that the
function π of π₯ has a discontinuity at π₯ equals three. Our answer to the question is,
therefore, false. The function represented by the
graph is not a continuous function. As a final point, we may note that
our function can have a discontinuity even though it is defined over all real
numbers.

Letβs now look at another graphical
example of determining whether a function is continuous or not.

Determine whether the function
represented by the graph is continuous or discontinuous.

For this question, weβve been given
a function π of π₯, which is defined when π₯ is greater than or equal to zero or
less than or equal to three. The interesting points of this
function occur when π₯ is equal to one and when π₯ is equal to two. Here, we see a sharp change in
gradient. And we may recognise that this
means our function would not be differentiable at these points.

For the purpose of continuity,
however, this is not necessarily cause for concern. Taking the point where π₯ equals
one as an example, we see that both the left- and the right-sided limits approach
the same value. And this will be where π of π₯ is
equal to one. From this, it follows that the
normal limit, as π₯ approaches one, will take the same value. And itβs also clear to see that π
of one is also equal to one.

In fact, these two things together
are the condition for continuity. Since we have that the normal
limit, as π₯ approaches one of π of π₯, is equal to π of one. The same logic would follow for the
point where π₯ is equal to two, and, in fact, for all other points along our
functionβs domain. This puts us in a position to
answer our question. We conclude that π of π₯ is a
continuous function.

Now, the two previous questions
gave us graphical examples of functions which were continuous and functions which
were not. In our example, we saw a function
with a jump discontinuity. However, functions with any other
discontinuity, such as removable or infinite, would also not be classed as
continuous functions. Conversely, there are many types of
functions which are continuous. And weβre gonna look at a few
examples algebraically of evaluating these.

The following types of functions
are continuous across their entire domain, polynomial functions, rational functions,
trigonometric functions, and exponential functions. An important distinction here is
weβre saying that these functions are continuous over their entire domain and not
for all values of π₯ over the real numbers. Weβll come back to this point
later. But before we do, another important
fact.

Sums, differences, products,
quotients, and compositions of continuous functions are also continuous for all
points where π₯ is properly defined. Again, these are within the domains
of our newly created functions. Now, the proof of continuity for
all of these types of functions is outside the scope of this video, but we can go
through one such example using a polynomial function and the following question.

What can be said about the
continuity of the function π₯ cubed plus five π₯ squared minus two π₯ plus two?

Given the general rule that a
polynomial function is continuous across its domain and the domain of our function
is all of the real numbers, we can almost immediately give the following answer. The function is continuous on β,
the real numbers, because it is a polynomial. This is, in fact, the quick answer
to our question.

But instead of just stopping here,
letβs explore a striped-back proof of why this is the case.

The building block that weβll start
with is the limit, as π₯ approaches π, of just some constant π. Now, clearly, the value of π₯
doesnβt matter to our constant π. And so, the answer to this limit is
just π. Here, weβll say that π is a real
number. Next, we move on to the case of the
limit, as π₯ approaches π, of the function which is just π₯. By taking a direct substitution
approach of π₯ equals π, we see that, here, the answer to our limit is just π. And also we should here stipulate
that π is also one of the real numbers.

Next, we add a power and we take
the limit as π₯ approaches π of π₯ raised to some power π. In this case, π is one of the
natural numbers, including zero. So, zero, one, two, three,
etcetera. With the same direct substitution
argument, we find that this limit is equal to π to the power of π. What about if we added some
constant π in front of our π₯ to the power of π? The constant multiple law allows us
to move our constant π outside of our limit as follows. Now, we see the limit weβre looking
for is the same as the previous line times π. And so, our answer is π times π
to the power of π, again with the direct substitution approach.

Okay, next, weβre gonna expand a
bit by adding two new terms to our limit and differentiating between different
values of π and π, where all values of π and π obey these same rules. Now, using the addition laws of
limits, we can split our one limit into three individual limits as follows. Now, you may notice that these
first two terms take the same form as the previous line of working that weβve just
completed. We can, therefore, say that they
are π one times π to the power of π one, and π two times π to the power of π
two, respectively. Of course, our last term is just a
constant, which we saw in our first example. The limit of this constant is, of
course, just π three.

We have now found that our limit is
π one times π to the power of π one plus π two times π to the power of π two
plus π three. Now, if we call the function that
weβve created here π of π₯, we should be able to see that the limit that we have
found is equal to our function evaluated at the point where π₯ is equal to π, in
other words, π of π.

Our final and crucial step is in
recognising the form of the function π of π₯ that weβve created. Since, π is a real number and π
is a natural number including zero, each of these terms gives us a real multiple of
π₯ raised to any positive integer power or zero. By the limit rule of addition, we
could add as many of these terms as we like to our function. Alongside this, weβve covered off
the addition of a standard real constant π. Our function π of π₯, therefore,
represents any polynomial that we care to construct.

Finally, weβve proved that the
limit, as π₯ approaches π, for our function π of π₯ is equal to π of π. Since π of π₯ is any polynomial
and π is any real number, we have just proved that a polynomial function is
continuous across the entire set of real numbers, since this is indeed the condition
for continuity.

Weβve now proved our continuity
condition for polynomials. And this proof can be extended or
altered to cover the other functions mentioned earlier. Now, when classifying that these
functions were continuous, we were careful to make the distinction that theyβre
continuous across their domain, and not for the entirety of the real numbers. As it happens, the domain of a
polynomial is the real numbers, however, this is not necessarily the case for other
functions such as rationals. Let us look at one such example to
illustrate this.

Find the set on which π of π₯
which is equal to π₯ minus 22 all over π₯ squared minus two π₯ minus 63 is
continuous.

For this question, we have a
rational function in the form π of π₯ over π of π₯. Now, we know that a rational
function is continuous over its domain. And hence, our question reduces to
finding the domain of our function π of π₯. In essence, we want to find values
of π₯ which would cause our function either to be undefined or to grow towards
positive or negative infinity. Looking at the form of our
function, we see that these troublesome points will occur when the denominator of
our quotient, π of π₯, evaluates to zero.

We can move forward with our
question first by factorising π of π₯. With a bit of inspection, we see
that this quadratic factorises to π₯ minus nine times π₯ plus seven since these two
numbers have a sum of negative two and a product of negative 63. From the factor theorem, we can
then see that when π₯ is nine or when π₯ is negative seven, π of π₯ will evaluate
to zero. Putting this factorised version of
π of π₯ back into our function, we can then conclude that π₯ equals nine and π₯
equals negative seven are not in the domain of our function. This is because, at these values,
the denominator of our quotient would be zero. And, hence, π of π₯ would not give
us a numerical evaluation.

Since π of π₯ behaves at all other
real values of π₯, we can say the following. The domain of our function π of π₯
is the real numbers minus the set of nine and negative seven. And so, π of π₯ is continuous over
the real numbers minus the set of nine and negative seven. And here, we have answered our
question.

To expand upon our previous
question, you may recall that when a common factor can be cancelled on the top- and
bottom-half of a quotient, such as that forming a rational function, this would
correspond to a removable singularity on our graph. In this case, since the function is
undefined where π₯ equals π, clearly, we cannot meet the criteria for
continuity. In cases where common factors
cannot be cancelled on the top- and bottom-halves of our quotients, we would expect
to see asymptotes where the values of π of π₯ would approach positive or negative
infinity.

At these asymptotes, even if the
left- and the right-limits were to agree, such as in this case, although we may
express that the limit, as π₯ approaches π of π of π₯, is equal to infinity, this
is just a particular way of saying that the limit does not exist. And, in fact, the limit does not
exist at any of these asymptotes. Now, since we have places where our
limit does not exist, again, we cannot fulfill our criteria for continuity. And at these points, π of π₯ would
be discontinuous.

Moving on, some functions can be
defined piecewise behaving differently over different intervals. The same rules of continuity apply
within the intervals of our piecewise function, but we must be careful when
examining the boundary between the intervals. In order for continuity to be
maintained over the boundary, the end points of the two sections, or subfunctions,
must join up. This is best illustrated with an
example.

Suppose that π of π₯ is equal to
five sin of π₯ minus three over π₯ minus three if π₯ is less than three and five π₯
squared over nine if π₯ is greater than or equal to three. Find the set on which π is
continuous.

Here, we have a piecewise function
defined over two intervals. The boundary of our two intervals
occurs at the point where π₯ equals three. And hence, this is an important
point. For our first subfunction, we have
a trigonometric expression on the top-half of our quotient and a binomial on the
bottom-half. Since we know that trigonometric
functions and polynomials are continuous on their domains, and quotients of
continuous functions are also continuous on their domains, we conclude that this
subfunction is, therefore, also continuous on its domain.

Now, here, we need to be slightly
careful since, at values where π₯ equals three, this subfunction evaluates to zero
over zero, which is an indeterminate form. Lucky for us, π of π₯ is only
defined by this subfunction at values of π₯ which are strictly less than three, not
where π₯ is equal to three. Instead, when π₯ is greater than or
equal to three, π of π₯ is defined by five π₯ squared over nine.

Again, here, itβs worth noting that
this monomial is defined over and has a domain of all real numbers. This means that the domain of our
function π of π₯ is all the real numbers. But we should be careful not to
hastily jump to the conclusion that π of π₯ is also continuous over all the real
numbers. Instead, we must still check the
criteria for continuity at the boundary between our subfunctions, or when π₯ is
equal to three.

First, letβs find π of three by
substituting in to five π₯ squared over nine. This value is easy to compute. And we get an answer of five. Next, we need to check that our
normal limit exists and is also equal to five. If this is not the case, then weβll
have a discontinuity at π₯ equals three. And hence, our function will not be
continuous at this point.

To move forward, we first recognise
that, either side of π₯ equals three, our function is defined by two different
subfunctions. To find our left limit, or when π₯
approaches from the negative direction, we use our first subfunction. Here, weβve already shown that a
direct substitution of π₯ equals three leads us to an indeterminate form of zero
over zero, so weβll need to use a different approach. Instead, we use the rule that the
limit, as π₯ approaches zero of sin π₯ over π₯, is equal to one.

Now, our expression isnβt in this
form. And so, we perform some
manipulations first by taking a factor of five outside of our limit using the
constant multiple rule. Next, we perform a
π’-substitution. By setting π’ as π₯ minus three, we
get the following. Our limit becomes sin π’ over
π’. However, we mustnβt forget to
change the limit value itself. π’ plus three is equal to π₯. Hence, π’ plus three approaches
three from the negative direction, or π’ approaches zero from the negative
direction.

Looking at our rule, we know that
if the normal limit exists and is equal to one, then the left- and right-sided
limits also exist and are also equal to one. We can now use our rule to evaluate
that this limit is equal to one. Hence, the left-sided limit, as π₯
approaches three of π of π₯, is equal to five times one, which is, of course, just
five. Now, our right-sided limit is far
easier to evaluate. For this, since weβre approaching
π₯ equals three from the right, we take the limit using our other subfunction. Simply, by direct substitution, we
find that this limit is equal to five.

Since the left- and right-sided
limits both exist and are equal to the same value, we can, therefore, conclude that
the limit, as π₯ approaches three of π of π₯, is also equal to five. And earlier, youβll recall that we
found that π of three is also equal to five. Since the limit, as π₯ approaches
three of π of π₯, is equal to π of three, we conclude that π of π₯ is continuous
when π₯ is equal to three.

If you were to think about this
visually, this means that the two end points of our subfunctions will join up. Let us now recall that, earlier, we
concluded that π of π₯ was continuous over all of the real numbers aside from
three, which we had to check. And now that we have checked three,
weβre in a position to say that our function π of π₯ is continuous on all the real
numbers. And this is the answer to our
question.

To wrap things up, letβs go through
a few key points. A function is continuous over an
interval if it is continuous over all points within said interval. A function, which weβll call π of
π₯, is continuous at a point, letβs say where π₯ equals π, if the limit, as π₯
approaches π of π of π₯, is equal to π of π. Here, we note the implication,
firstly, that this limit exists and, secondly, that the function is defined when π₯
equals π.

Polynomial, rational,
trigonometric, and exponential functions are continuous over their domains. And here, we note that this is not
necessarily all real numbers. Further to this, sums, differences,
products, quotients, and compositions of continuous functions are also continuous at
points where π₯ is properly defined.

Discontinuities of a function can
often be found by looking for values of x which result in a division by zero. And these may be removable or
essential discontinuities. The boundary between the intervals
of piecewise functions should be checked to make sure that the ends of the
subfunctions join up. If the left- and the right-sided
limits do not agree at these points, then the result will be a jump
discontinuity. If the left- and the right-sided
limits do agree, and theyβre both equal to the function evaluated at this point,
then continuity will be maintained.