17–83. Limits Evaluate the following limits. Use l’Hôpital’s R...

Question

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.

lim_x→∞ (ln(3x + 5eˣ)) / (ln(7x + 3e²ˣ)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to apply Lipita's rule to find the limit of the following function as X approaches infinity, and we're given the function F of X is equal to the natural log of the quantity of 2 X plus 4 multiplied by e to the X in quantity, divided by the natural. Log of the quantity of 5 X plus 2 multiplied by E to the 3 X in quantity. We're given 4 possible choices as our answers. For choice A, we have 2 divided by 5. For choice B, we have 4 divided by 5. For choice C, we have 1 divided by 3, and for choice D, we have 2 divided by 3. Now, we're asked to find the limit of our function F of X as X approaches infinity, using Lopital's rule. So we're gonna be looking for the limit. As X approaches infinity of F of X. Which is equal to the limit. As X approaches infinity. Of the natural log of the quantity of 2 X plus 4 multiplied by E to the X in quantity, divided by the natural log of the quantity of 5 X plus 2 multiplied by E to the 3 X in quantity. And if we were to try to evaluate this um limit by direct substitution, this would be equal to infinity divided by infinity. So here we have an indeterminate form. So, in order to evaluate this limit, we'll need to apply Lopital's rule. So recall for Lopital's rule, that we'll need to take the derivative with respect to X of both our numerator and denominator, and then reevaluate the limit. In doing so, this is going to be equal to the limit, as X approaches infinity. Of the derivative with respect to X of the natural log of the quantity of 2 X plus 4 multiplied by e to the X in quantity, divided by the derivative with respect to X of the natural log of the quantity of 5 X. Plus 2 multiplied by E 3 X in quantity. So, taking the derivatives, this is going to be equal to the limit. As X approaches infinity. Of in the numerator we'll have the derivative with respect to X of the natural log of the quantity of 2 X plus 4 multiplied by e to the X in quantity, and recall that when I take the derivative of the natural log function, I get 1 divided by the argument. So I'll have 1 divided by the quantity of 2 X plus 4 multiplied by e to the X in quantity. And then that gets multiplied by the derivative of the argument, since we need to apply our chain rule. So I'll have the derivative with respect to X of the quantity of 2 X plus 4 multiplied by e to the X, and when I take that derivative, I will get the quantity of 2 + 4 multiplied by E to the X in quantity. And then this is divided by the derivative with respect to X of the natural log of the quantity of 5 X plus 2 multiplied by E to the 3 X. And so again, recall that we're taking a derivative of our natural log function, so we get 1 divided by the argument. We'll have 1 divided by the quantity of 5 X plus 2 multiplied by E to the 3 X in quantity. Then that gets multiplied by the derivative of the argument. So I'll have the derivative with respect to X of the quantity of 5 X plus 2 multiplied by E to the 3 X. I'm gonna take that derivative, I get the quantity of 5, plus 6 multiplied by E to the 3 X in quantity. So now what we want to do is basically rearrange this fraction, and then break this up into two limits. So doing the rearranging, this is going to be equal to, for our first limit, we'll have the limit as X approaches infinity of the quantity. Of 2 + 4 multiplied by E to the X. In quantity, divided by the quantity of 2 X plus 4 multiplied by E to the X. In quantity. So here we basically just have rewritten our numerator, but that gets multiplied by the limit. As X approaches infinity. Of the quantity Of 5 X plus 2 multiplied by E to the 3 X in quantity, divided by the quantity of 5 plus 6 multiplied by E to the 3 X. In quantity. So here we've basically just moved everything in in our denominator up to the numerator and just rewritten the fraction that way. So now what we want to do is try to evaluate this limit again by direct substitution, and if I were to do that, this would be equal to for our first limit, we'd have infinity divided by infinity. That gets multiplied by in evaluating the second limit, we'd get infinity divided by infinity. So both of these limits give us an indeterminate form. So we'll apply Lopital's rule, uh, Lopital's rule again. So doing so, this is going to be equal to. The limit as X approaches infinity. Of the quantity of the derivative with respect to x. Of the quantity of 2 + 4 multiplied by E to the X in quantity, divided by the derivative with respect to X. Of the quantity of 2 X plus 4 multiplied by E to the X in quantity. That is multiplied by the limit. As X approaches infinity. Of the quantity of the derivative with respect to X of the quantity of 5 X plus 2 multiplied by E raised to the 3 X in quantity divided by the derivative with respect to X of the quantity of 5 plus 6 multiplied by E to the 3 X in quantity. So evaluating the derivatives, this is going to be equal to the limit as x approaches infinity of the quantity, and when I take the derivative with respect to X of the quantity of 2 + 4 multiplied by e to the X, I get 4 multiplied by E X, and that is divided by the derivative with respect to X of the quantity of 2 X plus 4 multiplied by e to the X, which gives us the quantity of 2 + 4 multiplied by e to the X in quantity. And this is multiplied by the limit. As x approaches infinity of the quantity, and I will have the derivative with respect to X of the quantity of 5 X plus 2 multiplied by E to the 3 X, and taking that derivative, I will get 5 plus 6 multiplied by E to the 3 X in quantity. And then that gets divided by the derivative with respect to X of the quantity of 5 plus 6 multiplied by E to the 3 X. And when I take that derivative, I get 18, multiplied by E to the 3 X. So now what we can do is simplify our expressions here for each of our limits. So, this is going to be equal to the limit, as X approaches infinity. Of the quantity, and here for our first term, I'm going to divide both the numerator and denominator by ETD X. So I will have 4. Divided by the quantity of 2 divided by E to the X. Plus 4 in quantity. That gets multiplied by the limit as X approaches infinity of the quantity, and here for our second term, I'm gonna divide both the numerator and denominator by E to the 3 X. So I'll have 5 divided by E to the 3 X. Plus 6, and that whole quantity is divided by 18. So now we can evaluate the limit by direct substitution. So this would be equal to, for our first term, we're going to have 4 divided by the quantity of 2 divided by infinity, which is 0, plus 4 in quantity. That gets multiplied by the quantity for our second um limit here, we will have 5 divided by infinity, which is 0. Plus 6, and that whole quantity is divided by 18. So, simplifying this expression, this is going to be equal to, we'll have 4 divided by 4, which is 1, and then we'll have 1 multiplied by 6 divided by 18, which is 1 divided by 3. So this here is the answer that we were looking for. And we can compare our answer to our answer choices. We see that the correct answer choice corresponds to answer choice C. So, I hope that this explanation has been useful, and I'll see you in the next video.

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.

lim_x→∞ (ln(3x + 5eˣ)) / (ln(7x + 3e²ˣ)

Key Concepts

Limits

L'Hôpital's Rule

Natural Logarithm

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