Does L'Hopital's Rule Apply to This Limit Problem?

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Discussion Overview

The discussion revolves around the application of L'Hôpital's Rule to the limit problem involving the expression \(\lim_{t\rightarrow 0}\ \frac{\ln t}{t^2-1}\). Participants explore whether the limit exists and the conditions under which L'Hôpital's Rule can be applied.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the limit, initially stating that it does not exist and struggling to apply L'Hôpital's Rule due to the form \(\frac{-\infty}{-1}\).
  • Another participant suggests that the limit should be evaluated as \(t\) approaches 1 instead of 0, indicating that L'Hôpital's Rule could then be applied to yield a limit of \(1/2\).
  • A third participant acknowledges the original wording of the question but does not dispute the proposed correction regarding the limit point.
  • One participant asserts that, as stated, L'Hôpital's Rule does not apply because the numerator approaches infinity while the denominator approaches 1, leading to the conclusion that the limit does not exist.
  • Another participant argues that the expression \((- \infty)/(-1)\) is not an indeterminate form and concludes that the limit can be computed directly, resulting in positive infinity.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the application of L'Hôpital's Rule or the existence of the limit. Multiple competing views are presented regarding the correct approach and interpretation of the limit problem.

Contextual Notes

There is ambiguity regarding the limit point (0 vs. 1) and the conditions under which L'Hôpital's Rule is applicable. The discussion reflects differing interpretations of the limit's behavior based on these factors.

faust9
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Howdy, I'm stumped by a seemingly easy l'hospital's limit question.

[tex]\lim_{t\rightarrow 0}\ \frac{\ln t}{t^2-1}[/tex]

I said the limit doesn't exist, but the asker claims it does. I tried various transformations so that I could use l'hospital's theorem but with no success. I keep getting [tex]\frac{- \infty}{-1}[/tex] so I can't take the derivative of the top and the bottom to get the answer.

Any help would be greatly appreciated.
 
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I think that the problem statement has a minor(?) error, it should be limit for t->1, not 0. Then L'Hospital's can be used (to get 1/2).
 
That's what I thought too, but the question was worded as I stated it...

Anyway, thanks for the input.
 
As stated, L'Hopital's rule does not apply: the numerator goes to infinity and the denominator goes to 1 so the limit does not exist.

The limits as x->1 or as x-> infinity both exist and can be done by L'Hopital. As mathman said, the limit at 1 is 1/2. The limit at
infinity is 0.
 
(-infinity)/(-1) is not an indeterminate form...it equals positive infinity

so the limit can be computed by simply plugging zero and then getting positive infintity
 

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