### Video Transcript

Limits of a Difference of
Powers

In this video, we will discuss and
prove several different results to help us evaluate the limits of a difference of
powers. We will also see several different
examples and applications of these results. Before we start with the general
case of the limit of a difference of powers, letβs start with a case weβve seen
before, the limit of a rational function. We recall if π of π₯ divided by π
of π₯ is a rational function β that means that both π and π are polynomials β then
we can evaluate the limit of our rational function by using direct substitution. The limit as π₯ approaches π of π
of π₯ over π of π₯ is equal to π of π divided by π of π. And this is, of course, provided
the denominator π of π is not equal to zero.

We can prove this directly from the
properties of limits. We just use the quotient rule for
limits and the fact that we can evaluate polynomials by direct substitution. And the rational function is an
example of a difference of powers. For example, π₯ to the πth power
divided by π₯ to the πth power is equal to π₯ to the power of π minus π. Weβll want to generalize this even
further. But for now letβs focus on the case
we have with rational functions. And weβll focus on the condition
that π of π is not allowed to be equal to zero.

To see how we might get around this
condition, letβs start by recalling the definition of a limit. We recall that we say the limit of
some function π of π₯ as π₯ approaches π is equal to some finite value of πΏ if
the values of π of π₯ approach πΏ as the values of π₯ approach π from both
sides. In particular, weβre only
interested in the output values of our function π of π₯ as our values of π₯
approach π. In other words, we want to know
what happens as our values of π₯ get closer and closer to π. We donβt actually mind what happens
when π₯ is equal to π.

We can use this to create a very
useful theorem. What if we had a function π of π₯
which was exactly equal to our function π of π₯ everywhere except when π₯ was equal
to π? So weβll start with a function π
of π₯, which is exactly equal to our function π of π₯ everywhere except when π₯ is
equal to π. We want to use this to determine
the limit as π₯ approaches π of π of π₯. And in fact we can do this. We know that π of π₯ is exactly
equal to π of π₯ everywhere except when π₯ is equal to π. And π of π₯ not being equal to π
of π₯ when π₯ is equal to π wonβt affect its limit because we donβt actually mind
what happens when π₯ is equal to π.

So in actual fact, their limits as
π₯ approach π must be equal. The limit as π₯ approaches π of π
of π₯ will be equal to the limit as π₯ approaches π of π of π₯. And this gives us a really useful
result. If we have two functions π of π₯
and π of π₯ which are equal everywhere except where π₯ is equal to π, and we know
the limit as π₯ approaches π of π of π₯ is equal to some finite value of πΏ, then
the limit as π₯ approaches π of π of π₯ must also be equal to πΏ.

Weβre now ready to apply this
result directly to our example involving rational functions. To do this, letβs start with an
example. Suppose we want to evaluate the
limit as π₯ approaches negative one of π₯ plus one multiplied by π₯ minus one all
divided by π₯ plus one. Since this is a rational function,
we can attempt to evaluate this limit by direct substitution. However, if we did this, in our
numerator we see π₯ plus one would become a factor of zero. And in our denominator, π₯ plus one
would also be equal to zero. So direct substitution gives us
zero divided by zero, which is an indeterminate form, which means we canβt evaluate
this limit by using direct substitution.

But itβs important to remember we
canβt just conclude the limit does not exist. All this tells us is that we canβt
evaluate this limit by using this method. We need to try a different
method. Instead, what we might try doing is
canceling the shared factor of π₯ plus one in the numerator and the denominator. This would then give us the limit
as π₯ approaches negative one of π₯ minus one, which we can evaluate by using direct
substitution. But we do need to be careful. Canceling the shared factor of π₯
plus one has changed the function weβre taking the limit of. Previously, negative one was not in
the domain of our function. However, negative one is in the
domain of π₯ minus one.

But we can justify that weβre
allowed to do this by using the property we just proved. When π₯ is not equal to negative
one, π₯ plus one is a nonzero number. And a nonzero number divided by
itself is always equal to one. What this really means is, our
original rational function and the polynomial π₯ minus one are equal everywhere
except when π₯ is equal to negative one. Therefore, by using our property,
their limits must be equal. And we can evaluate the limit of π₯
minus one as π₯ approaches negative one by using direct substitution. Itβs equal to negative one minus
one, which is of course just equal to negative two.

We can use this exact same
reasoning to evaluate the limit of other rational functions. So letβs clear some space and go
through one of these examples. We want to determine the limit as
π₯ approaches π of π₯ to the πth power minus π to the πth power all divided by
π₯ minus π. And for now, weβll assume that our
value of π is a positive integer. And this is a rational function, so
we could try evaluating this limit by direct substitution. If we did this, we would get π to
the πth power minus π to the πth power divided by π minus π, which simplifies
to give us zero divided by zero, which is of course an indeterminate form.

So instead, to evaluate this limit,
we want to use the same trick we did before. We want to cancel a shared factor
of π₯ minus π in the numerator and denominator. To do this, weβll start by calling
the polynomial in the numerator π of π₯. Thatβs π₯ to the πth power minus
π to the πth power. In particular, since π evaluated
at π is equal to zero, the remainder theorem tells us that π₯ minus π must be a
factor of π of π₯. In fact, by using polynomial
division or otherwise, we can show π₯ to the πth power minus π to the πth power
is equal to π₯ minus π all multiplied by π₯ to the power of π minus one plus π
times π₯ to the power of π minus two plus π squared multiplied by π₯ to the power
of π minus three. And we add terms of this form all
the way up to π to the power of π minus one.

We can then substitute this
expression directly into our limit. This then gives us the following
expression for our limit. And now we can cancel the shared
factor of π₯ minus π in both the numerator and denominator. And itβs worth reiterating weβre
allowed to do this because weβre taking the limit as π₯ approaches π. Canceling the shared factor of π₯
minus π will not change the value of our function anywhere except when π₯ is equal
to π. So we now have the limit as π₯
approaches π of π₯ to the power of π minus one plus π times π₯ to the power of π
minus two. And we add terms of this form all
the way up to π to the power of π minus one. And this is now the limit of a
polynomial, so we can evaluate this limit by using direct substitution.

This gives us π to the power of π
minus one plus π times π to the power of π minus two. And we add terms of this form all
the way up to π to the power of π minus one. And if we were to simplify each
term, we would notice something interesting. Every single term in this
expression is equal to π to the power of π minus one. And there are π of these terms,
one for each exponent of π₯, all the way from π minus one to zero. So, in fact, this is just equal to
π multiplied by π to the power of π minus one.

And although we assumed our value
of π was a positive integer, this result is true for any value of π. We have for any real constants π
and π their limit as π₯ approaches π of π₯ to the πth power minus π to the πth
power all divided by π₯ minus 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. Letβs now see an example of
applying this formula to evaluate a limit.

Find the limit as π₯ approaches one
of the fourth root of π₯ minus one multiplied by the sixth root of π₯ to the seventh
power minus one all divided by π₯ minus one all squared.

In this question, weβre asked to
evaluate a limit. And we can see that this is the
limit of a very complicated function. However, we can see that this
function is the sum, difference, quotient, product, and composition of power
functions and polynomials. So we can try evaluating this limit
by using direct substitution. If we substitute π₯ is equal to one
into this function and then simplify, we see that itβs equal to zero divided by
zero, which is an indeterminate form, which means we canβt evaluate the limit by
using this method. We need to try a different method
to evaluate this limit. Instead, we need to notice that the
limit weβre asked to evaluate is very similar to one of our limit results.

We know, for any real constants π
and π, the limit as π₯ approaches π of π₯ to the πth power minus π to the πth
power divided by π₯ minus π is equal to π times π to the power of π minus
one. And thatβs provided that both π to
the πth power and π to the power of π minus one exist. So we need to rewrite the given
limit in this form. To do this, weβll start by using
the product rule for limits to distribute the denominator over each of the factors
in our numerator. First, weβll rewrite our limit as
the limit as π₯ approaches one of the fourth root of π₯ minus one all divided by π₯
minus one multiplied by the sixth root of π₯ to the seventh power minus one all
divided by π₯ minus one.

Each of the two factors of our
function are now in the form of our limit rule. And by using the product rule for
limits, we can split the limit of a product of two functions into the product of the
limit of those two functions. And itβs worth reiterating this
will only be true provided the limit of both of our two functions exist. In fact, weβll be able to show this
by using our limit result. Before we apply this result, we
need to rewrite our numerator. Weβll rewrite the fourth root of π₯
by using our laws of exponents as π₯ to the power of one-quarter and the sixth root
of π₯ to the seventh power as π₯ to the power of seven over six.

Weβre now ready to use our limit
result to evaluate our limit. Letβs start with the first
limit. We have the value of π equal to
one-quarter and the value of π equal to one. Itβs worth noting that one raised
to the power of one-quarter is just equal to one. So this is in fact in the form of
our limit result. Therefore, by our limit result, we
can evaluate this limit to be equal to π multiplied by π to the power of π minus
one, which in this case is one-quarter multiplied by one raised to the power of
one-quarter minus one.

We can do exactly the same for our
second limit. The value of π is seven-sixths,
and the value of π is also equal to one. And once again, by our limit
result, we can evaluate this limit. Itβs π times π to the power of π
minus one, which in this case is seven-sixths multiplied by one raised to the power
of seven-sixths minus one. And of course we need to multiply
these two values together. And now we can evaluate this
expression directly. First, one raised to the power of
any number is just equal to one. So this simplifies to give us
one-quarter multiplied by seven-sixths, which is just equal to seven over 24.

Therefore, we were able to show the
limit as π₯ approaches one of the fourth root of π₯ minus one multiplied by the
sixth root of π₯ to the seventh power minus one all divided by π₯ minus one all
squared is equal to seven divided by 24.

Using our limit result for a
difference of two powers, we can actually show two other really useful limit
results. First, imagine we were asked to
evaluate the limit as π₯ approaches π of π₯ to the πth power minus π to the πth
power divided by π₯ to the πth power minus π to the πth power. We can actually write this entirely
in terms of our limit result. To do this, weβll start by
introducing a factor of π₯ minus π into both the numerator and denominator. Next, instead of multiplying, weβre
going to divide by the reciprocal. This gives us the following
expression. And we can see that both of these
two functions are in the form of our limit result. So weβre going to evaluate this
limit by using the quotient rule for limits.

The quotient rule for limits tells
us the limit of the quotient of two functions is equal to the quotient of the limits
of those two functions. Thatβs provided both of the two
limits exist and the limit in the denominator is not equal to zero. We can now evaluate both of these
limits by using our limit result. The first limit is equal to π
multiplied by π to the power of π minus one. And the second limit is equal to π
multiplied by π to the power of π minus one. We just need to divide both of
these two expressions. And when we divide these two
expressions and simplify, we get π over π multiplied by π to the power of π
minus π.

And this gives us a really useful
result. For any real constants π, π, and
π, the limit as π₯ approaches π of π₯ to the πth power minus π to the πth power
all divided by π₯ to the πth power minus π to the πth power is equal to π
divided by π multiplied by π to the power of π minus π. And thatβs provided π is not equal
to zero and π to the πth power, π to the πth power, and π to the power of π
minus π all exist.

Thereβs one more useful limit
result we can show from this. Weβre going to substitute π¦ is
equal to π₯ minus π into this limit result. To do this, letβs clear some space
and start with our limit result. We can find an expression for π₯ by
adding π to both sides. We see that π₯ is equal to π¦ plus
π. We can also see as the values of π₯
approach π, π₯ minus π is going to approach zero. So the values of π¦ will approach
zero. So by substituting π¦ is equal to
π₯ minus π into our limit, we get 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¦. And this is equal to π multiplied
by π to the power of π minus one. We can then rewrite this limit
result in terms of the variable π₯.

This gives us the following
result. For any real constants π and π,
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 both π to the
πth power and π to the power of π minus one exist.

Letβs now see an example of
applying one of these limit results.

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.

Letβs now go over some of the key
points we found in this lesson. First, if we have two functions π
of π₯ and π of π₯ which are equal everywhere except when π₯ is equal to π, and we
know the limit as π₯ approaches π of π of π₯ is equal to πΏ, then the limit as π₯
approaches π of π of π₯ must also be equal to πΏ. This is a really useful result. One thing it allows us to do is
cancel shared factors of π₯ minus π when weβre evaluating the limit of rational
functions.

We also showed three useful limit
results which hold for any real constants π, π, and π. First, the limit as π₯ approaches
π of π₯ to the πth power minus π to the πth power all divided by π₯ minus 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. Second, we showed the limit as π₯
approaches π of π₯ to the πth power minus π to the πth power all divided by π₯
to the πth power minus π to the πth power is equal to π divided by π multiplied
by π to the power of π minus π. And thatβs provided π is nonzero
and π to the πth power, π to the πth power, and π to the power of π minus π
all exist. And our final limit result told us
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.