### Video Transcript

In this video, weβre going to learn
about continuity at a point. This is a step on the way to
understanding continuous functions, functions like polynomial functions, exponential
functions, and certain trigonometric functions. Whose graphs can be drawn in a
single stroke of the pen. You may have noticed that, for a
lot of functions π, the limit of π of π₯ as π₯ approaches some value π is just π
evaluated at π, π of π. A function π is said to be
continuous at the point π if this happens. If the limit of π of π₯ as π₯
approaches π is just π of π. This is the definition of
continuity at a point. We can see then that the function
π is continuous at the point π if direct substitution works to find the limit of
π of π₯ as π₯ approaches π.

To understand continuity at a
point, we need to understand how a function can fail to be continuous at the point
π. How this equation can fail to hold
true. There are several things that can
go wrong. For example, this limit on the
left-hand side could not exist. For π to be continuous at π then,
we need this limit to exist. Similarly, we need the right-hand
side of the equation to exist. π must be defined at π. If it isnβt, then the right-hand
side of our equation, thatβs π of π, is undefined. And so our equation canβt possibly
hold. Another way of saying that π must
be defined at π is to say that π must be in the domain of π. What else could go wrong? Well, the limit on the left-hand
side could exist. And the function π of π could be
defined. But these values could be
different. We need the values of the limit of
π of π₯ as π₯ approaches π and π of π to be equal.

These are the three things we
needed to check to show that a function π is continuous at a number π. Letβs now apply this checklist to
some examples. For convenience, weβll always check
that π is defined at π before checking that the limit as π of π₯ approaches π
exists. And so the checklist for our
examples has a slightly different order, which Iβd encourage you to use too.

Given π of π₯ equals π₯ squared
plus π₯ minus two all over π₯ minus one, if possible or necessary, define π of one
so that π is continuous at π₯ equals one.

So we have a function π, which is
a rational function. And weβd like π to be continuous
at the point π₯ equals one. And weβre told that we should
define π of one to make this so, but only if it is possible or necessary. What does that mean? Well, if π is already continuous
at the point π₯ equals one, then it isnβt necessary to define π of one to make this
so. Itβs already been done for us. On the other hand, π could be
discontinuous at the point π₯ equals one. But in the way that itβs not
possible to things just by defining π of one. There could be some bigger issue
stopping π from being continuous at this point.

Letβs first check whether it is
necessary to define π of one to make π continuous at π₯ equals one. Is π already continuous at π₯
equals one? Well, we have a checklist which
allows us to find out whether a function is continuous at a certain point. π must be defined at that
point. So in our case, we need to check
that π of one is defined. And the limit of π of π₯ as π₯
approaches that point, in our case one, must exist. And finally, these two values must
be equal.

Letβs start at the top of the
list. Is π of one defined? Well, weβll use the definition of
π of π₯ that weβre given in the question. Substituting one for π₯ gives one
squared plus one minus two all over one minus one. In the numerator, one squared plus
one is two. And subtracting the two gives us
zero. And in the denominator, one minus
one is zero. So we get the indeterminate form
zero over zero. Zero over zero and hence π of one
are not defined. And so our function π is not
already continuous at π₯ equals one.

This isnβt necessarily a massive
issue. After all, our task is to define π
of one. If this is the only thing stopping
π from being continuous at π₯ equals one, then we can just define π of one. And π will be continuous at this
point, as required. We need to check then that there
arenβt any other issues. We need the limit of π of π₯ as π₯
approaches one to exist. And does it? We use the definition of π of π₯
from the question. And of course, we know that direct
substitution is going to give us an indeterminate form. So there must be some other way to
evaluate this limit.

Well, the factor theorem tells us
that as both the numerator and the denominator are zero when π₯ is one, both
numerator and denominator must have a factor of π₯ minus one. And indeed, we can factor the
numerator to π₯ plus two times π₯ minus one. Doing so allows us to cancel the
common factor of π₯ minus one in the numerator and denominator. To get the limit of π₯ plus two as
π₯ approaches one. This is a limit that can be
evaluated using direct substitution. Substituting one for π₯, we get one
plus two, which is three. So yes, the limit of π of π₯ as π₯
approaches one does exist. And it equals three.

Now the last thing on our checklist
is that the limit of π of π₯ as π₯ approaches one must be π of one. We found that the left-hand side,
the limit of π of π₯ as π₯ approaches one, is three. But the right-hand side π of one
is undefined. But our task is to define π of
one. If we define π of one to be three,
thatβs the value of the limit of π of π₯ as π₯ approaches one. Then our third item on the
checklist will be satisfied. And of course, defining π of one
to be three also sorts out the first item on our checklist. π of one is now defined. So defining π of one to be three
makes π continuous at π₯ equals one. Because π of one is now defined,
the limit of π of π₯ as π₯ approaches one, as we saw, is three. And now that weβve defined π of
one to also be three, these two values are equal.

It might be helpful to look at the
graph of π of π₯ to see what weβve done. As we saw with π₯ not equal to one,
π of π₯ is just π₯ plus two. And so the graph of π of π₯ is
just the straight line graph of π¦ equals π₯ plus two, with this hole here when π₯
is one. This hole in the graph comes
because π of one is not defined. And as a result, π is not
continuous at π₯ equals one. There is then this link between the
technical definition of continuity and our intuitive understanding of what
continuity means. There should be no gaps. We plug this gap and make the
function continuous by defining π of one to be three. Now, thereβs no hole in the
graph. And π is continuous at π₯ equals
one.

Letβs now see an example where we
canβt simply plug the gap.

Given π of π₯ equals one over π₯,
if possible or necessary, define π of zero so that π is continuous at π₯ equals
zero.

So we have the reciprocal function
here, which hopefully we know and love. And we know what its graph looks
like. The graph has two pieces, one in
the first quadrant and the other in the third. And these pieces are separated by a
vertical asymptote at π₯ equals zero. Intuitively then, it feels like
this function is not continuous at π₯ equals zero. Because the graph of our function
is not one continuous curve but is formed of two curves with the break at that
asymptote π₯ equals zero. Just to the left of this line when
π₯ is small but negative, π of π₯ is larger magnitude but negative. And to the right of this line when
π₯ is small but positive, π of π₯ is larger magnitude and positive. So as we pass π₯ equals zero, the
right of π of π₯ changes from a value which is larger magnitude and negative to a
value which is larger magnitude and positive.

Letβs see if our intuition agrees
with the technical definition by going through our checklist to see if the function
is continuous. For π to be continuous at π₯
equals zero, we need π of zero to be defined. The limit of π of π₯ as π₯
approaches zero to exist. And finally, we need these values
to be equal. Okay, so is π defined at zero? Well, if we substitute zero into
the definition of π of π₯, π of π₯ is one over π₯, we get π of zero is one over
zero. And this is not defined. Zero is not in the domain of our
function. So π is not defined at zero. And hence, π is not continuous at
π₯ equals zero. But this isnβt necessarily a
surprise as our task is to define π of zero to make π continuous at π₯ equals
zero. We wouldnβt have any work to do if
π were continuous at π₯ equals zero and π of zero was defined.

We check the second criterion. The limit of π of π₯ as π₯
approaches zero must exist. With the idea that if this limit
does exist, then we can just define π of zero to be the value of this limit. And, then the function will be
continuous at π₯ equals zero as required. Does this element exist? Well, if we look at the left-hand
limit as π₯ approaches zero, π of π₯ approaches negative infinity. And it gets worse. The limit from the right is
positive infinity. As π₯ approaches zero from the
right, π of π₯ gets larger and larger without bound.

The limit of π of π₯ as π₯
approaches zero therefore does not exist. And this is a problem. We can define π of zero to be
whatever we like. But whatever we define it to be,
this doesnβt change the fact that the limit of π of π₯ as π₯ approaches zero does
not exist. And so π cannot be continuous at
π₯ equals zero. So this is our answer then. The limit of π of π₯ as π₯
approaches zero doesnβt exist. And so π cannot be made continuous
at π₯ equals zero by defining π of zero. The fact that π of zero is
undefined is not really the problem here. And looking at the graph, we can
kind of see why this is true. There isnβt just a small gap on the
graph, which can be plugged by a single point. There is a huge chasm. While π of π₯ is not continuous at
π₯ equals zero and cannot be made so just by defining π of zero, itβs worth
pointing out that the function π of π₯ is actually continuous at other values of
π₯.

Given π of π₯ equals one over π₯,
if possible or necessary, define π of one so that π is continuous at π₯ equals
one. We go through our checklist with
this new question. We need π of one to be
defined. The limit of π of π₯ as π₯
approaches one to exist. And the limit of π of π₯ as π₯
approaches one to equal π of one. Is π of one defined? Yes, π of one is one over one,
which is one. π of one is defined. One is in the domain of π. Now, does the limit of π of π₯ as
π₯ approaches one exist? Letβs look at the graph. As π₯ approaches one from the left
and as π₯ approaches one from the right. The value of π of π₯ approaches
the same at finite value, the value one. If we wanted to find the value of
this limit without using the graph, weβd probably just use direct substitution. We can see then that the third item
on the checklist at the limit of π of π₯ as π₯ approaches one equals π of one is
also satisfied. The values of the left-hand side
and right-hand side are both one. Our answer then is that π is
already defined and continuous at π₯ equals one. Itβs not necessary to define π of
one again to make this so.

Letβs see one more quick example
before we conclude.

Is the function π of π₯, which is
piecewise defined to be two π₯ plus four all over π₯ plus two if π₯ is less than
negative two. Zero if π₯ is equal to negative
two. And π₯ squared plus six π₯ plus
eight all over π₯ plus two if π₯ is greater than negative two, continuous at π₯
equals negative two.

To find out, we go through our
checklist. For the function π to be
continuous at π₯ equals negative two, π of negative two must be defined. The limit of π of π₯ as π₯
approaches negative two must exist. And these values must be equal. So we start by checking that π of
negative two is defined. Well, thatβs easy to see from the
question. Weβre told that π of π₯ is zero if
π₯ is negative two. So π of negative two is
defined. π of negative two is equal to
zero. Now, we need to check that the
limit of π of π₯ as π₯ approaches negative two exists. There are different rules for the
values of π of π₯ when π₯ is less than negative two and when π₯ is greater than
negative two. Which weβll use to check the
left-hand and right-hand limits, making sure that they exist and are equal.

Whatβs the left-hand limit? Well, to the left of negative two,
π of π₯ is defined by two π₯ plus four all over π₯ plus two. So we have to find the limit of two
π₯ plus four all over π₯ plus two, as π₯ approaches negative two from the left. If we try direct substitution, we
get the indeterminate form zero over zero. So we have to find this limit some
other way. We do this by factoring the
numerator, getting two times π₯ plus two. And this factor of π₯ plus two
cancels with the factor of π₯ plus two in the denominator. And so our limit is that of the
constant function two whose value must be two. So the left-hand limit exists. What about the right-hand one?

To the right of π₯ equals negative
two, π of π₯ is this algebraic fraction. And we can find the value of this
limit in a similar way. Factoring the numerator and
canceling the common factor of π₯ plus two, to get the limit of π₯ plus four as π₯
approaches negative two from the right. This can be solved using direct
substitution. We get negative two plus four,
which is two. So this right-hand limit also
exists and is equal to the left-hand limit. Both of them have the value
two. Hence the limit of π of π₯ as π₯
approaches negative two from either direction also exists and has the value two.

All thatβs left to check is this
third point. For π of π₯ to be continuous at π₯
equals negative two, we need the limit of π of π₯ as π₯ approaches negative two to
be equal to the value of π at negative two, π of negative two. But weβve just seen that the value
of this limit is two, whereas π evaluated at negative two is zero. The limit of π of π₯ as π₯
approaches negative two is not equal to π of negative two. Although the first two conditions
on our checklist are satisfied, the last one isnβt. And so our answer is no. The function π is not continuous
at π₯ equals negative two, as the limit of π of π₯ as π₯ approaches negative two is
not equal to π of negative two.

Letβs now conclude with the key
points that weβve covered in this video. A function π is said to be
continuous at a number π if the limit of π of π₯ as π₯ approaches π equals π of
π. To check the continuity at π, we
must check that π of π is defined. That is, that π is in the domain
of π and that the limit of π of π₯ as π₯ approaches π exists. And only then does it make sense to
ask if they are equal. So if π of π is not defined or if
the limit doesnβt exist, then π is not continuous at π. However, if π of π is undefined,
but the limit of π of π₯ as π₯ approaches π exists, we can define π of π to be
the limit of π of π₯ as π₯ approaches π to make π continuous at π. If the limit of π of π₯ as π₯
approaches π does not exist, we canβt make π continuous at π by defining or
redefining π of π.