# Video: Finding the Limit at Infinity of the Quotient of Exponential Functions

Find lim_(π₯ β β) (2π^(3π₯) β 5)/(3π^(3π₯) β 1).

02:25

### 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. Weβll 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 itβs equal to positive infinity over positive infinity. And this is undefined. Therefore, weβve not yet found a solution.

However, the fact that it is equal to positive infinity over positive infinity does tell us that weβre able to use LβHopitalβs rule. LβHopitalβs 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 weβre taking the limit as π₯ tends to positive infinity. Therefore, weβre allowed to use LβHopitalβs 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 thereβs no π₯ dependency within our limit, our limit is simply equal to two-thirds. And this is the solution to the question.