Limit of x^x^x as x->0 from right

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Discussion Overview

The discussion revolves around evaluating the limit of the expression \( x^{x^{x}} \) as \( x \) approaches 0 from the right. Participants explore various methods for calculating this limit, including logarithmic transformations and the properties of exponential functions.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant notes their understanding of the limit \( \lim_{x \to 0^+} x^x \) and provides a calculation leading to 0.
  • Another participant suggests applying the same method to \( x^{x^{x}} \) but does not provide a complete solution.
  • A participant expresses difficulty in resolving the limit involving \( \ln x \) and \( x^x \), indicating they reached an indeterminate form.
  • There is a discussion about the indeterminate form \( -\infty \cdot 0 \) and its implications for the limit calculation.
  • One participant asserts that the limit should be 1, correcting an earlier assumption that it was 0, and emphasizes that \( \lim_{x \to 0^+} x^x = 1 \).
  • A later reply acknowledges a mistake in their earlier calculations, indicating confusion over the limit of \( x^x \) and its implications for the overall limit.

Areas of Agreement / Disagreement

Participants express differing views on the limit of \( x^{x^{x}} \) as \( x \) approaches 0 from the right, with some asserting it approaches 0 and others claiming it approaches 1. The discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Participants note that the limit involves indeterminate forms and that assumptions about the limits of \( x^x \) are critical to the discussion. There are indications of confusion regarding the application of logarithmic limits and the behavior of the expressions involved.

emilkh
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Any ideas?
lim XXX
x-> 0+

I know how to do x^x:
lim XX = lim ex * ln x = e0 = 0
lim x * ln x = lim ln x / (1/x) = lim (1/x) / (-1/x2) = lim -x = 0
 
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Now x^x is the exponent and x is the base, so do exactly as you just explained.
 
Could you elaborate more? I alredy tried this method and got stuck with
lim (ln x) * (x^x), i could not solve it
 
emilkh said:
Could you elaborate more? I alredy tried this method and got stuck with
lim (ln x) * (x^x), i could not solve it
Why are you stuck? You already said you knew what to do with x^x...
 
lim ln x * (x^x) = lim ln x * lim x ^ x = (- infinity) * 0 = 0 as X -> 0+

So you want to tell me that lim (X^x^x) = lim ex^x *ln x = e^0 = 1?

The limit suppose to be 0, .000001 ^ ( .000001 ^ .000001 ) = very very very small number (checked with calculator)
 
emilkh said:
lim ln x * (x^x) = lim ln x * lim x ^ x = (- infinity) * 0 = 0 as X -> 0+

Be careful; - \infty \cdot 0 is indeterminate.

Edit: I see, nevermind.
 
Last edited:
mutton said:
Be careful; - \infty \cdot 0 is indeterminate.
Well great! This is where I am stuck.

lim ln x * x was solved by switching it to lim ln x / (1/x) and taking derivatives, with lim ln x * x/x it's not going to work
 
emilkh said:
Any ideas?
lim XXX
x-> 0+

I know how to do x^x:
lim XX = lim ex * ln x = e0 = 0
lim x * ln x = lim ln x / (1/x) = lim (1/x) / (-1/x2) = lim -x = 0

e0 is not =0, but...1..... here is your mistake ,

hence ...limx^x=1 as x tends to 0 from the right

And lim (ln x) * lim x^x = infinity multiplied by 1 and NOT by 0
 
[edited for content], in my solutions it was 1, but somehow I copied formula wrong and the whole time assumes lim x^x = 0. I spend way too much time studying for finals... got to take break
 
Last edited by a moderator:

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