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$\lim_{x\to 0 } \frac{x}{\ln x}.$

This was wrong, I got a big red wrong! Why doesn't L'Hôpital work on this one? The problem is that $\ln$ is not defined for 0. It needs to be rewritten?

(Thanks to everyone helping me out with my homework, due anxiety I'm not able attend the class workshops.)

Edit: I did get the correct limit ($0$), but that was a coincidence.

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    Maybe I just have to explicitly say what the domai$n$ of the function is positive numbers, and than apply L'Hopital? I've heard that a function isn't a function until it's domain have been mention :p2010-11-21

4 Answers 4

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L'Hopital's rule is best forgotten about. Of course as $x\to0^+$, $\log x\to-\infty$ and so $1/\log x\to0$. A fortiori $x/\log x\to 0$.

A better problem is to find $\lim_{x\to0^+}x\log x$.

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    @Algific, Robin is not doing any sort of domain restirction; he is noting that the logarithm "has no floor" as one goes nearer and nearer the origin; thus its reciprocal should approach zero.2010-11-21
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The limit, as written, does not exist because the function is not defined on an open neighborhood of the limit point: the function is not defined on $(-\infty,0)$, so you cannot take the limit as $x\to 0$. That would be a full answer.

If, however, the limit is taken only from the right, then the other answers you have received are correct: the limit is not an indeterminate form, it is equal to $0$. But you cannot get that by 'evaluating', because $\ln(x)$ is not defined at $0$.

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    I really should stop using wolfram. THank!2010-11-21
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$\frac{0}{\infty}$ is not indeterminate form, is zero, then $\lim\limits_{ x \rightarrow 0^+ }\frac{x}{lnx}=0$ but is a right limit.

But L'Hôpital's rule applies..

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    L'Hopital's Rule *does not apply*, at least if by LR you mean what is found in any American calculus text I have ever seen or this wikipedia article: http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule. In any case, as a sometime calculus teacher it is distressing to see someone evaluate this limit by citing some fancy theorem: a calculus student should be able to reason that if the numerator of a fraction is geting very small and the denominator is getting very large, then the entire value of the fraction gets arbitrarily small.2011-07-16
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As others have mentioned, you don't need L'Hospital's rule for this simple example. However, in fact, there is a more general form of L'Hôpital's rule that requires only that the denominator $\to\infty\:.$ This handles your case and many others so it is well-worth knowing. For an exposition see the papers below and see the theorem in Rudin's PoMA excerpted in my answer here.

A. E. Taylor, $\ $ L'Hospital's Rule
Amer. Math. Monthly, Vol. 59, No. 1 (Jan., 1952), pp. 20-24.
http://www.jstor.org/stable/2307183

A. M. Ostrowski, $\ $ Note on the Bernoulli-L'Hospital Rule
Amer. Math. Monthly, Vol. 83, No. 4 (Apr., 1976), pp. 239-242.
http://www.jstor.org/stable/2318210

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    @Arturo: Yes,$I$agree - it depends upon contextual definitions.2010-11-21