### Video Transcript

Find the limit as π₯ tends to
infinity of two timesed by π to the power of three π₯ minus five over three times
π to the power of three π₯ minus one.

We can start by trying to find this
limit using direct substitution. Weβll be using the fact that the
limit as π₯ tends to infinity of π to the power of π₯ is equal to positive
infinity. From this, we obtain that the limit
as π₯ tends to infinity of π to the power of three π₯ is also equal to positive
infinity. And this tells us that when we use
direct substitution in order to find our limit, we find that itβs equal to positive
infinity over positive infinity. And this is undefined. Therefore, weβve not yet found a
solution.

However, the fact that it is equal
to positive infinity over positive infinity does tell us that weβre able to use
LβHopitalβs rule. LβHopitalβs rule tells us that if
the limit as π₯ approaches π of π of π₯ over π of π₯ is equal to zero over zero,
positive infinity over positive infinity, or negative infinity over negative
infinity. Where π is a real number, positive
infinity, or negative infinity. Then the limit as π₯ approaches π
of π of π₯ over π of π₯ is equal to the limit as π₯ approaches π of π prime of
π₯ over π prime of π₯.

Now our limit is equal to infinity
over infinity. And weβre taking the limit as π₯
tends to positive infinity. Therefore, weβre allowed to use
LβHopitalβs rule. In our case, π of π₯ is equal to
two times π to the power of three π₯ minus five. And π of π₯ is equal to three
timesed by π to the power of three π₯ minus one.

We find π prime and π prime by
differentiating π and π. Differentiating two π to the power
of three π₯ minus five with respect to π₯, we obtain that π prime of π₯ must be
equal to six π to the power of three π₯. And differentiating three times π
to the power of three π₯ minus one with respect to π₯, we obtain that π prime of π₯
must be equal to nine π to the three π₯. We obtain that our limit must be
equal to the limit as π₯ tends to infinity of six timesed by π to the three π₯ over
nine timesed by π to the three π₯.

Here we notice that we have a
factor of three timesed by π to the three π₯ in both the numerator and
denominator. Since we can write our numerator as
two timesed by three π to the power of three π₯ and our denominator as three
timesed by three π to the power of three π₯. Therefore, these factors of three
timesed by π to the power of three π₯ will cancel out, leaving us with the limit as
π₯ tends to infinity of two over three. Since thereβs no π₯ dependency
within our limit, our limit is simply equal to two-thirds. And this is the solution to the
question.