### Video Transcript

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.