### Video Transcript

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

In this question, weβre asked to
evaluate the limit of a function. We can see in our numerator we have
a polynomial and in our denominator we have a polynomial. So this is a rational function. And we could always try to evaluate
the limit of a rational function by direct substitution. Substituting π₯ is equal to two
into the function, we get two minus four all cubed plus eight all divided by two
minus two, which, if we evaluate, we see is zero divided by zero, which is an
indeterminate form. Since this gives an indeterminate
form, we canβt evaluate this limit by using direct substitution; weβre going to need
to use a different method.

We need to notice the limit given
to us in the question is very similar to one of our limit results. That is, the limit as π₯ approaches
zero of π₯ plus π all raised to the πth power minus π to the πth power all
divided by π₯ is equal to π times π to the power of π minus one. And thatβs provided π to the πth
power and π to the power of π minus one both exist. But this limit result has π₯
approaching zero, and the limit weβre asked to evaluate has π₯ approaching two. So weβre going to use the
substitution π¦ is equal to π₯ minus two. Then, as our values of π₯ approach
two, π₯ minus two is going to be approaching zero. So our values of π¦ approach
zero.

In our denominator, we have π₯
minus two, which is just going to be equal to π¦. However, in our numerator, we have
π₯ minus four. So we need to find an expression
for π₯ minus four. And we can find this by subtracting
two from both sides of our equation for π¦. We get π¦ minus two is equal to π₯
minus four. Therefore, by using the
substitution π¦ is equal to π₯ minus two, we were able to rewrite our limit as the
limit as π¦ approaches zero of π¦ minus two all cubed plus eight all divided by
π¦.

And this is now almost exactly in
the form of our limit result. We can write it in the exact form
of our limit result by noting that π¦ plus negative two is the same as π¦ minus two
and eight is the same as negative one times negative two all cubed. So our value of π is negative two
and our value of π is three. Therefore, our limit result tells
us that this limit is equal to π multiplied by π to the power of π minus one. Substituting π is equal to
negative two and π is equal to three, we get three multiplied by negative two to
the power of three minus one, which we can evaluate is equal to 12. Therefore, we were able to show the
limit as π₯ approaches two of π₯ minus four all cubed plus eight all divided by π₯
minus two is equal to 12.