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Why doesn't L'Hôpital's rule work here? $$\lim_{x\to0}\frac{x^2}{x^2-x}$$ By L'Hôpital's rule I have $1$ while the limit is $0$....

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By l'hipital you get 2x/(2x-1) -> 0.. Why do you say it goes to 1? – Ant Jun 20 '14 at 8:39
@user147985 It goes to zero only ... try again – Debashish Jun 20 '14 at 8:44
It does work. One of the hypothesis is that the limit you're computing is undetermined if taken naively. But after you apply L'Hôpital once, you can already get the limit naively, it's not $\frac 00$ anymore, you can'use the theorem again. So $\lim \limits_{x\to 0}\left(\dfrac{x^2}{x^2-x}\right)=\lim \limits_{x\to 0}\left(\dfrac{2x}{2x-1}\right)=\lim \limits_{x\to 0}\left(\dfrac{2}{2}\right)=1$ is wrong. – Git Gud Jun 20 '14 at 8:45
@ Git Gud Thank you very much! – user147985 Jun 20 '14 at 8:47
You can also cancel out a factor of $x$ from the numerator and denominator to remove the discontinuity at 0. – Hao Ye Jun 20 '14 at 9:32

$$\lim_{x\rightarrow 0} \frac{2x}{2x-1} = \frac{\lim_{x\rightarrow 0}2x}{\lim_{x\rightarrow 0}2x-1} = \frac{0}{-1} = 0 \ne 1$$

For the completeness of my answer:
You can't use L'Hôpital the second time, since this rule demands both numerator and denominator converges to $0$ or diverges to infinity. Instead, evaluate seperatedly the limit for the numerator and denominator (arithmetic of limits) as I demonstrated above.

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But when I do L'Hôpital's rule again I have lim 2/2=1... – user147985 Jun 20 '14 at 8:45
You can't use L'Hôpital's rule the second time (the denominator isn't $0$). – AnnieOK Jun 20 '14 at 8:48
You don't have the necessary conditions to use the rule again, since the limit doesn't present an inditerminate form. – Victor Jun 20 '14 at 8:55
The necessary conditions are: f(x)->0, g(x)->0, and f'(x)/g'(x) has a limit. The second of these conditions is not met. – gnasher729 Jun 20 '14 at 9:23
@AnnieOK why not include that in your answer? As it is now, it looks a bit snarky. – Matsemann Jun 20 '14 at 14:54

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