The discussion revolves around evaluating the limit of \(x^x\) as \(x\) approaches zero, particularly in the context of understanding the behavior of the integral \(\int_0^e \ln(x)\). Participants explore whether this limit diverges and how it relates to the integral's convergence.
There is an ongoing exploration of different methods to evaluate the limit, with some participants questioning the relevance of \(x^x\) in the context of the integral. Guidance is offered regarding the application of l'Hôpital's Rule, and various interpretations of the limit are being discussed without reaching a consensus.
Participants are navigating the complexities of limits and integrals, with some expressing uncertainty about the mathematical processes involved. The discussion reflects a mix of intuitive reasoning and formal approaches, highlighting the challenges of evaluating limits in this context.
Nathanael said:Homework Statement
I want to integrate \int_0^e \ln(x) but first, I wondered if it would be divergent. I figured if xx goes to zero as x goes to zero then the integral would diverge (because xln(x)-x would diverge).
2. The attempt at a solution
I'm wondering how you could show that this limit (xx as x→0) is 1. It makes intuitive sense that it should approach 1, and it's clear from the graph, but I'm curious as to how someone would find this limit mathematically.
Ah, right. I forgot about l'Hopital's idea.Dick said:Write it in such a way that you can use l'Hopital.
Nathanael said:Ah, right. I forgot about l'Hopital's idea.
lim of \frac{x}{\ln(x)^{-1}} as x→0 = lim of \frac{1}{-(x\ln(x)^2)^{-1}} as x→0 = lim of -(x\ln(x)^2) as x→0
I feel like I could apply this indefinitely with no progress...
Ah, I see!
If
lim of x\ln(x) as x→0 = lim of -x\ln(x)^2 as x→0
then
lim of x\ln(x) as x→0 = lim of -x as x→0 = 0
Thanks Dick.
P.S.
Sorry if this was hard to read; I don't know how to write the "x→0" under the "lim" in tex. (If someone wants to teach me, it would be appreciated :) )
I thought l'Hopital's idea only applied if the numerator function and denominator function both approach zero? That's why I wrote it as x/(1/ln(x))Dick said:I'm not really sure what you are doing there. Try writing x*ln(x) as ln(x)/(1/x) and do l'Hopital.
L'Hopital's Rule can also be used when the limit has the form ##[\frac{\infty}{\infty}]##.Nathanael said:I thought l'Hopital's idea only applied if the numerator function and denominator function both approach zero? That's why I wrote it as x/(1/ln(x))
Nathanael said:If your way works too that's great (and perhaps a bit simpler) but I arrived at the answer in the previous post.
Nathanael said:I thought l'Hopital's idea only applied if the numerator function and denominator function both approach zero? That's why I wrote it as x/(1/ln(x))
If your way works too that's great (and perhaps a bit simpler) but I arrived at the answer in the previous post.
\lim\limits_{x\to 0}\big( x\ln(x)\big)=\lim\limits_{x\to 0}\big(-x\ln(x)^2\big)Dick said:I don't know why you just dropped the ln(x)^2.
Nathanael said:\lim\limits_{x\to 0}\big( x\ln(x)\big)=\lim\limits_{x\to 0}\big(-x\ln(x)^2\big)
I divided both sides by -ln(x)
\lim\limits_{x\to 0}\big(x\ln(x)\big)=\lim\limits_{x\to 0}(-x)=0
The reason I reversed the order was so it makes more sense reading it from left to right, but now I see that it's confusing. Sorry.
Nathanael said:Homework Statement
I want to integrate \int_0^e \ln(x) but first, I wondered if it would be divergent. I figured if xx goes to zero as x goes to zero then the integral would diverge (because xln(x)-x would diverge).
Right. I was just thinking about xx because I thought perhaps it would be easier to determine the limit of that than the limit of xln(x). It's easier to guess the limit of xx, but to actually calculate the limit we had to go back back to the original xln(x). So xx was indeed irrelevant.ehild said:You have shown that ln(xx)=x ln(x) goes to zero at the limit x=0. So xx goes to 1.
You do not need the limit of xx, as the integral is x(ln(x)-1) . And it is not divergent if x goes to zero.