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

Find lim_(𝑥 → ∞) (2𝑒^(3𝑥) − 5)/(3𝑒^(3𝑥) − 1).

02:25

### Video Transcript

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

We can start by trying to find this limit using direct substitution. We’ll be using the fact that the limit as 𝑥 tends to infinity of 𝑒 to the power of 𝑥 is equal to positive infinity. From this, we obtain that the limit as 𝑥 tends to infinity of 𝑒 to the power of three 𝑥 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 it’s equal to positive infinity over positive infinity. And this is undefined. Therefore, we’ve not yet found a solution.

However, the fact that it is equal to positive infinity over positive infinity does tell us that we’re able to use L’Hopital’s rule. L’Hopital’s rule tells us that if the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to zero over zero, positive infinity over positive infinity, or negative infinity over negative infinity. Where 𝑎 is a real number, positive infinity, or negative infinity. Then the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 over 𝑔 prime of 𝑥.

Now our limit is equal to infinity over infinity. And we’re taking the limit as 𝑥 tends to positive infinity. Therefore, we’re allowed to use L’Hopital’s rule. In our case, 𝑓 of 𝑥 is equal to two times 𝑒 to the power of three 𝑥 minus five. And 𝑔 of 𝑥 is equal to three timesed by 𝑒 to the power of three 𝑥 minus one.

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

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