Limit as x goes to zero of x^x

[Nathanael]

Homework Statement

I want to integrate \int_0^e \ln(x) but first, I wondered if it would be divergent. I figured if x^x 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 (x^x 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.

 

[Dick]

Homework Statement

I want to integrate \int_0^e \ln(x) but first, I wondered if it would be divergent. I figured if x^x 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 (x^x 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.

The log of x^x is ln(x)*x. Try and find the limit of that. Write it in such a way that you can use l'Hopital.

 

[Nathanael]

Write it in such a way that you can use l'Hopital.

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...

Oh, 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 :) )

 

[Dick]

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'm not really sure what you are doing there. Try writing x*ln(x) as ln(x)/(1/x) and do l'Hopital.

 

[Nathanael]

I'm not really sure what you are doing there. Try writing x*ln(x) as ln(x)/(1/x) and do l'Hopital.

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.

 

[Mark44]

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))

L'Hopital's Rule can also be used when the limit has the form $[\frac{\infty}{\infty}]$.

If your way works too that's great (and perhaps a bit simpler) but I arrived at the answer in the previous post.

 

[Dick]

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.

Not really. x*ln(x)^2 isn't any easier as a limit than x*ln(x). I don't know why you just dropped the ln(x)^2.

 

[Nathanael]

I don't know why you just dropped the ln(x)^2.

\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.

 

[Dick]

\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.

Now I see. That's an original way to do it. Look ok to me.

 

[ehild]

Homework Statement

I want to integrate \int_0^e \ln(x) but first, I wondered if it would be divergent. I figured if x^x goes to zero as x goes to zero then the integral would diverge (because xln(x)-x would diverge).

You have shown that ln(x^x)=x ln(x) goes to zero at the limit x=0. So x^x goes to 1.
You do not need the limit of x^x, as the integral is x(ln(x)-1) . And it is not divergent if x goes to zero.

 

[Nathanael]

You have shown that ln(x^x)=x ln(x) goes to zero at the limit x=0. So x^x goes to 1.
You do not need the limit of x^x, as the integral is x(ln(x)-1) . And it is not divergent if x goes to zero.

Right. I was just thinking about x^x 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 x^x, but to actually calculate the limit we had to go back back to the original xln(x). So x^x was indeed irrelevant.