Limit as x goes to zero of x^x

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In summary, we were trying to determine the limit of the function x*ln(x) as x approaches 0 in order to integrate \int_0^e \ln(x). However, we realized that we did not need to find this limit, as the integral can be written as x(ln(x)-1) and is not divergent as x goes to 0.
  • #1
Nathanael
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Homework Statement


I want to integrate [itex]\int_0^e \ln(x)[/itex] 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.
 
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  • #2
Nathanael said:

Homework Statement


I want to integrate [itex]\int_0^e \ln(x)[/itex] 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.

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.
 
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  • #3
Dick said:
Write it in such a way that you can use l'Hopital.
Ah, right. I forgot about l'Hopital's idea.

lim of [itex]\frac{x}{\ln(x)^{-1}}[/itex] as x→0 = lim of [itex]\frac{1}{-(x\ln(x)^2)^{-1}}[/itex] as x→0 = lim of [itex]-(x\ln(x)^2)[/itex] as x→0

I feel like I could apply this indefinitely with no progress...

Oh, I see...

If
lim of [itex]x\ln(x)[/itex] as x→0 = lim of [itex]-x\ln(x)^2[/itex] as x→0
then
lim of [itex]x\ln(x)[/itex] as x→0 = lim of [itex]-x[/itex] 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 :) )
 
  • #4
Nathanael said:
Ah, right. I forgot about l'Hopital's idea.

lim of [itex]\frac{x}{\ln(x)^{-1}}[/itex] as x→0 = lim of [itex]\frac{1}{-(x\ln(x)^2)^{-1}}[/itex] as x→0 = lim of [itex]-(x\ln(x)^2)[/itex] as x→0

I feel like I could apply this indefinitely with no progress...

Ah, I see!
If
lim of [itex]x\ln(x)[/itex] as x→0 = lim of [itex]-x\ln(x)^2[/itex] as x→0
then
lim of [itex]x\ln(x)[/itex] as x→0 = lim of [itex]-x[/itex] 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.
 
  • #5
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.
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.
 
  • #6
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))
L'Hopital's Rule can also be used when the limit has the form ##[\frac{\infty}{\infty}]##.
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.
 
  • #7
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.

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.
 
  • #8
Dick said:
I don't know why you just dropped the ln(x)^2.
[itex]\lim\limits_{x\to 0}\big( x\ln(x)\big)=\lim\limits_{x\to 0}\big(-x\ln(x)^2\big)[/itex]

I divided both sides by -ln(x)

[itex]\lim\limits_{x\to 0}\big(x\ln(x)\big)=\lim\limits_{x\to 0}(-x)=0[/itex]

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.
 
  • #9
Nathanael said:
[itex]\lim\limits_{x\to 0}\big( x\ln(x)\big)=\lim\limits_{x\to 0}\big(-x\ln(x)^2\big)[/itex]

I divided both sides by -ln(x)

[itex]\lim\limits_{x\to 0}\big(x\ln(x)\big)=\lim\limits_{x\to 0}(-x)=0[/itex]

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.
 
  • #10
Nathanael said:

Homework Statement


I want to integrate [itex]\int_0^e \ln(x)[/itex] 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).

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.
 
  • #11
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.
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.
 

1. What is the limit as x approaches zero of x^x?

The limit as x approaches zero of x^x is equal to 1.

2. How do you find the limit as x goes to zero of x^x?

To find the limit as x goes to zero of x^x, you can use L'Hopital's rule or rewrite the expression as e^(xlnx) and then use the limit definition of e as x approaches zero. Both methods will result in the limit being equal to 1.

3. Why is the limit as x approaches zero of x^x equal to 1?

The limit as x approaches zero of x^x is equal to 1 because as x gets closer and closer to zero, the value of x^x also gets closer and closer to 1. This is due to the fact that any number raised to the power of zero is equal to 1, and as x approaches zero, the value of x gets closer to 1.

4. Is the limit as x goes to zero of x^x a one-sided or two-sided limit?

The limit as x goes to zero of x^x is a two-sided limit, meaning it is evaluated from both the left and right sides of x=0. This is because the behavior of x^x as x approaches zero is different from the left and right sides.

5. Can the limit as x goes to zero of x^x be evaluated using direct substitution?

No, the limit as x goes to zero of x^x cannot be evaluated using direct substitution because it would result in an indeterminate form of 0^0. Instead, a different method such as L'Hopital's rule or the limit definition of e must be used to find the limit.

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