Help Solving: Euler's Limit Question (x > 0)

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Discussion Overview

The discussion revolves around the limit expression \(\lim_{t\rightarrow 0^{+}} \frac{1-x^t}{t} = -\ln x\) for \(x > 0\) as presented in William Dunham's book. Participants explore methods to derive this result, particularly focusing on the application of L'Hopital's rule.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant seeks clarification on how the limit results in \(-\ln x\).
  • Multiple participants suggest using L'Hopital's rule to evaluate the limit.
  • One participant acknowledges a misunderstanding in their initial approach, noting they differentiated with respect to \(x\) instead of \(t\).
  • Another participant humorously comments on the preference for \(t\)-differentiation in certain contexts, specifically in non-linear optics.
  • A later post outlines a step-by-step application of L'Hopital's rule, indicating that the limit evaluates to \(-\ln x\) by differentiating the numerator and denominator.
  • Another participant reiterates that the derivative of \(x^t\) with respect to \(t\) leads to the conclusion when evaluated at \(t=0\).

Areas of Agreement / Disagreement

Participants generally agree on the use of L'Hopital's rule as a method to solve the limit, but there is no consensus on the humor presented in the discussion or the clarity of the explanation regarding differentiation.

Contextual Notes

Some participants express confusion regarding the application of differentiation with respect to the correct variable, indicating a potential misunderstanding of the limit process. The discussion does not resolve the nuances of the humor or its relevance to the mathematical problem.

murshid_islam
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In the preface to William Dunham's book "Euler: The Master of Us All", he wrote,

[tex]\lim_{t\rightarrow 0^{+}} \frac{1-x^t}{t} = -\ln x[/tex] for [tex]x > 0[/tex]

I can't figure out how the result comes out to be [itex]-\ln x[/itex]. Can anyone help?
 
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Use L'Hopital's rule
 
arildno said:
Use L'Hopital's rule
Got it! Thanks!

I was trying the L'Hopital's rule before, but I was mistakenly trying it by differentiating with respect to x instead of t.
 
murshid_islam said:
Got it! Thanks!
You're welcome! :smile:

I was trying the L'Hopital's rule before, but I was mistakenly trying it by differentiating with respect to x instead of t.
In general, in particular in non-linear optics, x-differentiation should never be chosen if a t-differentiation is possible. :wink:
 
arildno said:
In general, in particular in non-linear optics, x-differentiation should never be chosen if a t-differentiation is possible. :wink:
Didn't get the joke. :confused:
 
It wasn't particularly funny, sort of an inductio-ad-aburdum, pythonesque sort of joke i often make, and that falls flat, SPLAT! on the ground...:cry:
 
hello welcome
i think it is so easy, right
we will apply L'Hopital's rule for top and bottom

top = 1 - x^t ==> -x^t lnx
bottom = t ==> 1

then limit t----->0+ -x^t lnx / 1
t=0
so our limit will be -lnx
hi
-x^t = -x^0 =1
 
its just minus the derivative of x^t wrt t, at t=0, so the basic rule that the derivative of x^t is ln(x).x^t gives it, for t = 0.
 

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