Can L'Hopital's Rule be applied to limits with multiple zeros?

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Discussion Overview

The discussion revolves around the application of L'Hopital's Rule to limits involving indeterminate forms, specifically the limit of (x^x)/((e^x)-1) as x approaches 0. Participants explore the nature of the limit and the conditions under which L'Hopital's Rule may be applicable.

Discussion Character

  • Exploratory, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant questions whether L'Hopital's Rule can be applied since the limit involves the form (0^0)/0, which may not fit the standard 0/0 form required for the rule.
  • Another participant suggests that e^x - 1 behaves like x near zero, leading to the expression approximating x^(1+x), which approaches 0.
  • A different viewpoint posits that if the approximation is correct, the limit would yield x^(x-1) approaching infinity.
  • One participant acknowledges a misunderstanding in their previous calculations, indicating the complexity of the limit evaluation.

Areas of Agreement / Disagreement

Participants express differing interpretations of the limit's behavior and the applicability of L'Hopital's Rule, indicating that multiple competing views remain without a consensus on the correct approach.

Contextual Notes

The discussion highlights uncertainties regarding the form of the limit and the assumptions made in approximating e^x - 1, as well as the implications of the indeterminate form involved.

rman144
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I have been working with a limit for a while now but cannot for the life of me seem to solve it. Any ideas:


lim[x appr. 0] of (x^x)/((e^x)-1)


I've tried turning x^x into e^(x ln(x)), but the root of my problem is that I'm unsure of whether or not I can use L'Hopital's because technically, (0^0)/0 is not necessarily of the form 0/0.
 
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e^x - 1 ~ x. Therefore your expression is ~ x^(1+x) -> 0^1 = 0.
 
You make a good point, but, assuming that is correct, wouldn't it go to:

x^(x-1) >>> oo
 
rman144 said:
You make a good point, but, assuming that is correct, wouldn't it go to:

x^(x-1) >>> oo

That is right.
This is straight forward
x^x->1
exp(x)-1->0
1/0->infinity
 
rman144 said:
You make a good point, but, assuming that is correct, wouldn't it go to:

x^(x-1) >>> oo
You're right. I misread your original expression - I missed the / .
 

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