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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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Why not, $lim_{x\to 0} \frac{1}{x} = \infty$, so after differentiation of numerator and denominator, you can take limits. – Timothy Wagner Nov 21 '10 at 16:52
Maybe I just have to explicitly say what the domain 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 :p – Algific Nov 21 '10 at 17:33

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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Is this a fully valid answer? Can I just restrict the domain of the function at my will, when no domain is mention? – Algific Nov 21 '10 at 17:40
Algific, what on earth are you talking about? – Robin Chapman Nov 21 '10 at 21:45
@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. – J. M. Nov 21 '10 at 23:10

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! – Algific Nov 21 '10 at 20:32

$\frac{0}{\infty}$ 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: 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. – Pete L. Clark Jul 16 '11 at 1:58

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.

A. M. Ostrowski, $\ $ Note on the Bernoulli-L'Hospital Rule
Amer. Math. Monthly, Vol. 83, No. 4 (Apr., 1976), pp. 239-242.

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But the x -> 0 not x->0+. I know that if indeed it said x->0+ I could have used L'hopital. – Algific Nov 21 '10 at 17:56
Both are the same when you're working with log as a real function. Presumably your answer was marked wrong because the course did not present this more general form of L'Hospital's rule (and possibly the grader did not know it). See the links above for the precise statement. – Bill Dubuque Nov 21 '10 at 19:01
from what I can tell (don't have full access from home), the first link is talking about limits along the interval $I$ which has $c$ as an extremity; so it would correspond here to the limit as $x\to 0^+$, rather than as $x\to 0$. Under some strict interpretations of "$\lim_{x\to 0}$", for $\log$ (and others, e.g., for $\sqrt{x}$), $\lim_{x\to 0}\log(x)$ does not exist, because the left handed limit (as $x\to 0^-$) cannot be evaluated. – Arturo Magidin Nov 21 '10 at 20:39
@Arturo: Yes, I agree - it depends upon contextual definitions. – Bill Dubuque Nov 21 '10 at 20:52

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