### Video Transcript

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.