### Video Transcript

Evaluating Limits Using Algebraic
Techniques.

In this video, weβll learn how to
use techniques such as factorisation to take the limit of a function. First, let us recall the definition
of a limit, which is, for some function π of π₯, which is defined near the value
where π₯ is equal to π, as π₯ approaches π, π of π₯ approaches πΏ. We call this value πΏ the
limit. And here, we have written the
standard notation for the description of a limit.

Now, if π of π₯ is a rational
function with π in its domain, we can simply say that the limit, as π₯ approaches
π of π of π₯, is equal to π of π. Since weβre inputting the value of
π into our function, this method is often described as direct substitution. An important point to note here is
that even if π is not in the domain of our function π, we can still sometimes find
the limit as π₯ approaches π. This is because the limit concerns
values as π₯ approaches π but is not equal to π. And weβll see how this comes into
play in a moment.

For this video, weβll be focusing
on functions which take the form π of π₯ over π of π₯ where both π and π are
polynomial functions. In the cases weβll look at, where
π₯ is equal to π, π of π over π of π will evaluate to the indeterminate form of
zero over zero. That is to say, if we try to take
the limit as π₯ approaches π of our function π of π₯ by directly substituting π₯
equals π into the function, then our substitution fails, and weβll need to take
another approach.

Looking at our quotient, we can see
that both π of π and π of π are equal to zero. By recalling the factor theorem, we
can then infer that both π of π₯ and π of π₯ will have a common factor of π₯ minus
π. Given this information, we can
rewrite the function uppercase π of π₯ as a product of π₯ minus π times some other
function, which weβll call lowercase π of π₯. And, of course, the same logic can
be followed for uppercase π of π₯.

This allows us to rewrite our
original quotient as π₯ minus π times lowercase π of π₯ divided by π₯ minus π
times lowercase π of π₯. In this form, we see that the
common factor of π₯ minus π can be cancelled in the top and bottom halves of our
quotient. And we are then left with lowercase
π of π₯ over lowercase π of π₯. Letβs define this new quotient as
π of π₯.

Now, a reasonable question to ask
is, why are we giving this new definition when π of π₯ is clearly equal to π of
π₯. In fact, the answer is that in
cancelling the common factor of π₯ minus π, we have changed the domain of our
function π of π₯. We, therefore, have that π of π₯
equals π of π₯, but not at the point where π₯ is equal to π. The subtlety here is that π of π₯
is defined when π₯ is equal to π, whereas π of π₯ is not.

This is great, as it allows us to
proceed to the following general rule. The limit, as π₯ approaches π of
π of π₯, is equal to the limit as π₯ approaches π of π of π₯ if π and π are
equal at all points over an interval except the point at which π₯ is equal to
π. Again, itβs good to understand that
this works because the limit concerns values of π₯ which are close to π but not
equal to π. Finally, since π is a rational
function with π in its domain, we can simply find the limit by direct substitution,
which is π evaluated where π₯ is equal to π. Okay, thatβs a lot of information,
so letβs look at an example to illustrate this concept.

Find the limit as π₯ approaches
negative eight of π₯ squared plus 13π₯ plus 40 divided by π₯ cubed plus nine π₯
squared minus 12π₯ minus 160.

Here, we see that we are trying to
find the limit of a rational function, which weβll call π of π₯. For this type of function, if
negative eight is in the domain of π, then the limit, as π₯ approaches negative
eight, is simply π evaluated at negative eight. The first thing we can do then is
to try direct substitution of negative eight into our function.

Here, we have performed the
substitution. And following through with our
working, we find that our answer evaluates to zero over zero. And this is an indeterminate form,
which means we cannot evaluate the limit using direct substitution. We can also conclude that negative
eight is not in the domain of our function π. Here, weβre gonna need to use a
different approach. And we can do so first by
recognising that our function π of π₯ is in the form π of π₯ over π of π₯, where
both π and π are polynomial functions.

Looking at the failed direct
substitution that we just tried, we can see the both π of negative eight and π of
negative eight are equal to zero. From this information, we can use
the factor theorem to conclude that π₯ plus eight is a factor of both π of π₯ and
π of π₯. Since direct substitution failed,
letβs instead try to factorise both the top and bottom halves of our quotient, given
the fact that both of them have a common factor of π₯ plus eight.

For the numerator, factoring a
quadratic equation should be a familiar skill to us. And with some inspection, we see
that π₯ squared plus 13π₯ plus 40 factorises to π₯ plus eight and π₯ plus five. Now, for the denominator,
ordinarily, factoring a cubic is a much more difficult task. However, given the fact that we
already know one of the factors is π₯ plus eight, we can use techniques such as
polynomial division or comparing coefficients to find the other factor.

For this video, weβre going to
compare coefficients. And we begin by recognising that
our other factor will take the form ππ₯ squared plus ππ₯ plus π. To find π, we note that we have an
π₯ multiplied by an ππ₯ squared. And this is equal to π₯ cubed. It, therefore, follows that π is
equal to one. And we then see that our second
factor begins with the term π₯ squared.

Next, to find π, we note that we
have eight multiplied by π is equal to negative 160. And we can calculate that π is
equal to negative 20. Finally, to find π, weβll choose
to look at the nine π₯ squared term. Going by the previous coefficients
we have, we note that we first have an eight multiplied by an π₯ squared. This gives us eight π₯ squared. We also have an π₯ multiplied by
ππ₯ and this gives us ππ₯ squared. The sum of these two terms is nine
π₯ squared. And it then follows that π is
equal to one.

The denominator of our quotient is
now fully factorised. And our missing factor was π₯
squared plus π₯ minus 20. Following this factorisation, weβre
able to cancel the common factor of π₯ plus eight from the top and bottom half of
our quotient. And we are then left with π₯ plus
five over π₯ squared plus π₯ minus 20.

Now, here, we should remember that
in cancelling the common factor of π₯ plus eight, we have changed the domain of our
original function π of π₯. Weβre able to say that π of π₯ is
equal to the right-hand side of our equation, which weβll call π of π₯, at all
values where π₯ is not equal to negative eight. It now follows that the limit, as
π₯ approaches negative eight of π of π₯, is equal to the limit as π₯ approaches
negative π₯ of π of π₯. Crucially, this is because the
limit concerns values where π₯ is close to negative eight but not equal to negative
eight.

An important point is that π of π₯
is defined when π₯ is equal to negative eight. And so, we can, therefore, find the
limit by direct substitution. Here, we have performed the
substitution of π of negative eight. And if we follow through with our
calculations, we see that this evaluates to negative three over 36. This fraction can be simplified to
negative one over 12. We have now answered the
question. And we have found our limit.

This example illustrates that when
our functions π of π₯ and π of π₯ evaluate to zero, we can use the factor theorem
to help us with potentially tricky factorisations such as cubics. Letβs now look at another technique
which can help us avoid potentially tricky factorisations first by focusing on
expressions of the form π₯ to the π minus π to the π, or the difference of two
πth powers.

Since inputting π₯ equals π into
this equation will always give zero, our friend the factor theorem, again, tells us
that all expressions of this form will evaluate to zero. And therefore, all of them will
have a factor of π₯ minus π. Using this general rule, we see
that our second factor will take the form of a polynomial with π terms having
decreasing powers of π₯ and increasing powers of π up to the powers of π minus
one. Some examples of this have been
shown below.

This general rule can be used to
derive a very useful formula in the following way. Consider the case of a difference
of two πth powers, π₯ to the π minus π to the π, divided by the difference of
two πth powers, π₯ to the π minus π to the π. We can use our general rule to
express the top and bottom halves of our quotient as a product of two factors. In both cases, one of these factors
is π₯ minus π.

We can cancel this common factor of
π₯ minus π in the top and bottom halves of the quotient. In cancelling our common factor, we
must remember that the left- and right-hand sides of our equations are equal as long
as the value of π₯ is not equal to π. Since the limit, as π₯ approaches
π, concerns values of π₯ close to π but not equal to π, weβre able to say that
the limit of the left-hand side is equal to the limit of the right-hand side.

Considering the numerator of our
quotient for a moment, we now use the following trick of direct substitution, where
π₯ takes the value of π. With this substitution, weβre left
with the sequence π terms long of which all of the terms are equal to π to the
power of π minus one. And this is simply equal to π
times π to the power of π minus one. Although, we were using the
numerator for an example, the same logic follows for the denominator of our quotient
and gives us π times π to the power of π minus one.

We then proceed with a few
simplifications of the powers of π. We then find that our limit, as π₯
approaches π, is equal to π over π multiplied by π to the power of π minus
π. This is a really useful result for
equations of this form since even when the powers π or π are large, we can avoid
lengthy factorisations when taking the limit. Letβs now see how this technique
can be used in an example.

Find the limit as π₯ approaches two
of eight π₯ cubed minus 64 divided by π₯ squared minus four.

Here, we have a function, which
weβll call π of π₯. Given that this is a rational
function, the first thing we may try is a direct substitution of π₯ equals two into
our function. Here, we have performed the
substitution. And when we evaluate our answer, we
find that weβre left with the indeterminate form of zero over zero. Instead, weβre gonna need to move
on to a different method based on factorisation.

For this method, we first note that
our function π of π₯ is in the form π of π₯ over π of π₯, where both π and π
are polynomial functions. By inspecting the numerator of our
quotient, we noticed that both the terms have a factor of eight. We can, therefore, factorise our
numerator as eight times π₯ cubed minus eight. And given that this eight is a
constant, we can take it outside of our limit as follows. If instead we find the limit as π₯
approaches two of π₯ cubed minus eight divided by π₯ squared minus four and multiply
this entire thing by our constant eight, this will give us the same answer.

To proceed, we can then notice that
the eight in our numerator and the four in our denominator can both be expressed as
powers of two, which are two cubed and two squared, respectively. After doing this, we see that our
limit now takes the following form. Here, weβll say that the top half
of our quotient is equal to the difference of two πth powers, with π being three,
and the bottom half of our quotient is equal to the difference of two πth powers,
with π being two.

Given this form, weβre able to use
the following general rule, which tells us that the limit will be equal to π over
π times π to the power of π minus π. At this point, you may notice that
both the top and bottom half of our quotient would have a common factor of π₯ minus
two. We could, instead, cancel this
common factor and proceeded by refactorising. However, this general rule allows
us to move directly to our limit.

In cases, where π or π are large,
this helps us potentially avoid a lengthy or time-consuming refactorisation. Inputting our values into our
general rule, where π is two, π is three, and π is also two, we find that our
limit is three over two multiplied by two to the power of three minus two. We also mustnβt forget to multiply
this entire thing by the eight which we took out of our limit.

Three minus two is, of course, just
one. And so, we can simplify this by
cancelling the two and the one over two. We then find that our answer is
equal to eight times three, which 24. We have now found that the limit,
as π₯ approaches two of our function π of π₯, is equal to 24. And we have answered our
question.

The general rule that we used in
this video provides us with a useful shortcut that can be used when our function π
of π₯ can be expressed in this form. Again, although for our question
the powers involved are relatively low, when π and π are sufficiently large, you
may be able to save yourself much more time. An interesting point worth noting
is that some of the techniques that we have shown in this video can also be used to
take the limits of functions which include radical. Specifically, where π of π₯ is
equal to π of π₯ over π of π₯, where either π or π include a radical term, such
as the square root of π₯.

Although these would not be classed
as rational functions, some of our tools will still work. An example would be if we were to
take the limit as π₯ approaches eight of the cube root of π₯ minus two divided by π₯
minus eight. If we were to try a direct
substitution approach, we would again be left with the indeterminate form of zero
over zero.

However, if instead we were to
factorise, specifically taking out a factor of the cube root of π₯ minus two from
our denominator, we would then be able to proceed in a familiar fashion. By cancelling out the common factor
in the top and bottom halves of the quotient and then directly substituting π₯
equals eight into the remaining expression to finally find an answer of one over
12. Although, here, weβve shown an
example where our current techniques work, in some cases, weβll need a different
method based on multiplication by a conjugate. This technique can be illustrated
using an example.

Determine the limit as π₯
approaches eight of π₯ squared minus eight π₯ divided by the square root of π₯ plus
one minus three.

Here, we see that weβre taking the
limit of a function π of π₯, which takes the form π of π₯ over π of π₯. When we try a direct substitution
approach to solve this, our expression evaluates to the indeterminate form of zero
over zero. And instead, weβre gonna need to
use a different approach. The first thing we can notice is
that our numerator can be factorised as π₯ times π₯ minus eight. Given our knowledge of the factor
theorem, we may have even expected this factor of π₯ minus eight given that the
function π evaluated where π₯ equals eight gave us a zero.

Unfortunately, the denominator of
our quotient is less straightforward to evaluate. But we can understand how to move
forward by viewing it as a binomial in the form π minus π, where π is equal to
the square root of π₯ minus one and π is equal to three. The conjugate of any binomial is
found by flipping the symbol in between the two terms. The conjugate of our binomial would
then be π plus π. And in this case, that is the
square root of π₯ plus one plus three.

Now, multiplying a binomial by its
conjugate is a useful tool because it leaves us with the difference of two
squares. Given that our denominator contains
a square root, we can utilise this relationship to evaluate our function. Now, multiplying by the conjugate
over itself is the same as multiplying by one. Nonetheless, the bottom half of our
quotient then simplifies to a difference of two squares.

And with some working, we see that
our denominator then becomes π₯ minus eight. Following this, weβre able to
cancel the common factor of π₯ minus eight in the top and bottom half of the
quotient. And on the right-hand side of our
equation, we are left with π₯ multiplied by the square root of π₯ plus one plus
three.

Now, since our original function π
of π₯ is equal to the right-hand side of our equation at all points where π₯ is not
equal to eight, weβre able to say the limit, as π₯ approaches eight of our original
function π of π₯, is equal to the limit as π₯ approaches eight of our new function,
which weβll call π of π₯. This is because the limit concerns
values of π₯ which are arbitrarily close to eight but not equal to eight.

In this form, we can perform a
direct substitution of π₯ equals eight into our function π to find the limit. Performing the substitution and
then evaluating our answer leaves us with a value of 48. And this is, in fact, the limit
that we were looking to find.

Now, itβs worth recapping here
that, originally, we were not able to find our limit using direct substitution
because we were left with the indeterminate form of zero over zero. Instead, after factorising and
using our conjugate method to cancel a common factor, direct substitution was
possible. We should, therefore, be on the
lookout to use this method for questions of this form when we observe a radical in
our quotient.

Letβs now recap a few key points to
summarise. When taking the limit of a rational
function π of π₯, which is expressed as π of π₯ over π of π₯, and evaluating by
direct substitution, we may be left with the indeterminate form of zero over
zero. In these cases, if the function π
of π₯ is almost equal to π of π₯. That is to say π of π₯ is equal to
π of π₯ at all values of π₯ aside from where π₯ is equal to π. Then it follows that the limit, as
π₯ approaches π of π of π₯, is equal to the limit as π₯ approaches π of π of
π₯.

Assuming that π is in the domain
of our new function π, that is to say π of π₯ is defined where π₯ is equal to π,
we can then find our limit by direct substitution, which is to say taking the value
of π of π. We can find such a function π of
π₯ via a number of methods such as factorisation or multiplication by a conjugate of
the numerator or denominator.

The methods that we have shown in
this video involve cancelling a common factor of π₯ minus π from the top and bottom
half of our quotient. Since taking the limit involves
values of π which are close to but not equal to π, it does not matter that in
cancelling the common factor we have changed the domain of our original function π
of π₯. Finally, when our function takes
certain forms, there are some useful shortcuts that we can take to find the limit
directly, potentially avoiding some lengthy factorisations.