Use l’Hôpital’s rule to find the limits in Exercises 7–52.
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Question

Use l’Hôpital’s rule to find the limits in Exercises 7–52.

32. lim (x → 0) (3^x - 1) / (2^x - 1)

Accepted Answer

Welcome back, everyone. Find the value of the following limit using Lapital's rule. Limit as x approaches 0 of 7 to the power of X minus 1 divided by 3 of X minus 1. He says LN of 3 divided by LN of 7. B LN of 7 minus LN of 3. C LN of 7 divided by LN of 3. In D the limit does not exist. So for this problem, let's rewrite the limit. Limit as X approach is 0. Of 7 to the power of x minus 1. Divided by 3 to the power of X minus 1. Let's check if lapel's rule is applicable. We're going to substitute X equals 0, which is 7 to the power of 0, that's 1 minus 1 divided by. We to the power of 0 gives us 1 minus 1, right? So we get 0 divided by 0. Which is an indeterminate form and it works for Lapito's rule, meaning we can immediately apply the rule, and it says that we want to differentiate the numerator and the denominator. So we want to take the derivative of 7 to the power of X minus 1. And divided by the derivative of treats the power of x minus 1. Let's find each derivative. For the numerator, the derivative of 7 to the power of X is 7 X LN of 7. The derivative of a constant is 0. And in the denominator, we're going to get 3 to the power of X and N of 3. Once again, the derivative of -1 is going to be 0. And now we are trying to substitute X equals 0 again, right? So we get 7. So the power of 0, which is 1, right? Multiplied by LN of 7, so basically 1 multiplied by LN of 7. Divided by 3 to the power of 0, which is 1. Which is multiplied by LN of 3. So now we get a finite result, right? LN of 7. Divided by a line of 3. Therefore, the correct answer for this problem corresponds to the answer choice C. Thank you for watching.

Use l’Hôpital’s rule to find the limits in Exercises 7–52.32. lim (x → 0) (3^x - 1) / (2^x - 1)

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Use l’Hôpital’s rule to find the limits in Exercises 7–52.
32. lim (x → 0) (3^x - 1) / (2^x - 1)