### Video Transcript

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.