evaluate-each-limit-if-it-exists-use-l-hospitals-rule-if-appropriate-lim-_x-rightarrow-infty-frac1e2

Question

Evaluate each limit (if it exists). Use $L$ Hospital's rule (if appropriate).
$$\lim _{x \rightarrow \infty} \frac{1+e^{2 x}}{2+\ln x}$$

Gregory Higby · Accepted Answer

So we&#x27;re looking at the limit as X goes to infinity of one plus E. To the two X over two plus natural log effects. And uh if you look at the graphs of those two functions, like shifting up doesn&#x27;t really matter. Um And the expo doesn&#x27;t really matter. So that&#x27;s the numerator, it&#x27;s definitely going I think it&#x27;s definite that it&#x27;s going to infinity if you look at the graph. Uh and then same thing with two plus natural log. So it just shifts the graph up. But the denominator is still going to infinity as well. So we have this indeterminant form of infinity over infinity to do low petals rule. Um Yeah. So how do we do love details rules? So he leaves the limit the same and we take the derivative of the top. So the heat of the two X. Because the derivative of one is zero. And then don&#x27;t forget the chain rule. You have to take the derivative of two X. Which is to over the derivative of natural log of X would be one over X. Um So as we&#x27;re looking at this problem, the numerator is still going to infinity. Going infinitely up. But the graph of one over X may not do this in a different color as we move to the right, we&#x27;re getting closer and closer to zero. So this is no longer indeterminate form. Um But as far as the uh well maybe I shouldn&#x27;t think of it this way. Maybe what I should be thinking about is instead of dividing by one over X, we can multiply by the reciprocal. So that function would be the same thing as two X. E. To the two X. Yeah, this is a better way of thinking about it because now we&#x27;re taking infinity and multiplying by another infinity. So we are going infinitely up. I&#x27;m just not a fan of saying it equals infinity because that&#x27;s not a number and limits need to equal a number. You might have a teacher that&#x27;s okay with that as your answer, but I would rather see does not exist Danny, but this is correct as well.

Carson Merrill · Answer

We&#x27;re trying to find the limit as X approaches infinity of the function. And as normal, we are going to want to take the natural log of both sides to put it in a simpler form. That simpler form is going to be the natural log of two times the natural log of X divided by one plus the natural log of X. Then when we evaluate this at infinity we get infinity over infinity. This is an indeterminate form. So we&#x27;re going to want to use low tells rule when we do this in the numerator will get Natural log of two times one over X. And then in the denominator we will just get one over X. So if that one over acts in one of our X will cancel out giving us just natural log of two as a result. Um So now that we know that&#x27;s natural log of two, we also know that the natural log of why? Because we had to take the natural log of both sides. So we have that the natural log of Y equals the natural log of two. Based on that, it&#x27;s clear that why is going to be equal to two? So chew is our final answer.

Amy Jiang · Answer

Okay, so we want to find our fallen women. Good. Start for using drugs, though. So we can plug in Exeter Cathedral celebrity or to third party three times minus Naja. Log 00 plus one. So this gives us that we have to do cars. RL minus the natural log of one. So this is equal to one minus. They&#x27;re all What? Did you quit to one?

Amy Jiang · Answer

I guess you want to find the limit of the following equation. So why don&#x27;t we use our limit long lived than to put this up into our numerator and denominator that as we get the limit of our numerator? And then that&#x27;s over the limits of our, um, for the nominee? I don&#x27;t hear that we have X squared plus specs so we can talk her out of X. So looking canceled its X, I note that X cannot be able to go, so that gives us Let&#x27;s try using drugs off. We have e negative to time zero plus one over a zero plus one. Cities gives us each a part of 1/1, so we see that our limits, but that&#x27;s just evil to eat.

Jonathan Kerby-White · Answer

here we have the limits as X approaches zero from positive side one, minus Ellen of X over one plus Ellen FX. So we just, uh we use direct substitution here to get one from the positive side. Well, vexed the graph looks way that says we approach zero x axis y axis. It&#x27;s a rough sketch of it. Natural. Lagerback&#x27;s is it approaches zero From the positive side, we go to negative infinity. So over one plus negative infinity. So here we just scared infinity over negative infinity So we can use Loki does rule So here we applaud low Pato&#x27;s rule So limited X ghosts zero From the positive side, we take the derivative over one, we get zero minus the derivative of the natural log of X negative one over x the same way about him. This equals the limit. This X approach zero from positive side negative one. So c equals negative one

Carson Merrill · Answer

So here we&#x27;re going to be finding the limit um possibly through using hotels rule, we have to determine that. So we see that in this case we have X to the X. Right, that&#x27;s one. And since we&#x27;re approaching zero, that&#x27;s going to be a potential issue. Um And then we&#x27;ll have this divided by the natural log of X. Let&#x27;s act my next one. That&#x27;s what we&#x27;re dealing with here. And we want to know the limit as X approaches zero from the right. So in this case we wouldn&#x27;t really use locals rule instead. What we can do is take the limit as X goes to zero from the right, um and recognize what&#x27;s happening. So looking at the graph here, we see that as we approach X from the right or as we approach zero from the right, It&#x27;s constantly approaching a value of zero. So based on that, we see that the answer is going to be zero.

Problem

Evaluate each limit (if it exists). Use $L$ Hospi…

Problem 23 Easy Difficulty

Evaluate each limit (if it exists). Use $L$ Hospital's rule (if appropriate).
$$\lim _{x \rightarrow \infty} \frac{1+e^{2 x}}{2+\ln x}$$

Answer

$\infty$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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Basic Technical Mathematics with Calculus

Grace H.

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

Catherine R.

Michael J.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus