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I have trouble finding the limit of the following : $$ \lim\limits_{x\to 1} \frac{(\ln x)^2}{1-x} $$ using the rule from L´Hopital.

Since both quotients converge to $0$, I should be able to use L´Hopitals´s rule right?

But when I do that the derivations converge to $$\frac{0}{-1}$$ Does that mean that there's no limit?

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    $\begingroup$ No it doesn't. It means just that: the limit is $\frac{0}{-1}=0$ $\endgroup$ – SBareS Dec 21 '14 at 14:55
  • $\begingroup$ So it just means that it converges to $0$ ? thats it ? $\endgroup$ – ViktorG Dec 21 '14 at 14:56
  • $\begingroup$ Yep, that's it. $\endgroup$ – SBareS Dec 21 '14 at 14:56
  • $\begingroup$ It's spelled "L'Hopital." $\endgroup$ – fluffy Dec 21 '14 at 21:27
  • $\begingroup$ @fluffy my bad, changed it $\endgroup$ – ViktorG Dec 21 '14 at 21:29
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Using l'Hopital's rule, we find $$ \lim_{x \to 1} \frac {\ln^2 x}{1-x} = \lim_{x \to 1} \frac {2 \ln x \cdot \frac {1}{x}}{-1} = \lim_{x \to 1} - \frac {2 \ln x}{x} = \lim_{x \to 1} 0 = \boxed {0}. $$We know l'Hopital's rule applies because it is of the form $ \frac {\ln^2 1}{1-1} = \frac {0}{0} $, which is indeed indeterminate.

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  • $\begingroup$ But $2ln(x)\cdot\frac{1}{x}$ converges to $0$ no? So you have $\frac{0}{-1}$ and that is $0$. That´s at least how i got it. Maybe I´m wrong ... $\endgroup$ – ViktorG Dec 21 '14 at 15:08
  • $\begingroup$ @DavidMitra Yes, you are right. I made a mistake, but I've fixed it. Were you the one who downvoted? If so, I suggest you revert. :) $\endgroup$ – Ahaan S. Rungta Dec 21 '14 at 15:12
  • $\begingroup$ @ViktorG Did you downvote? I suggest you revert now. :) $\endgroup$ – Ahaan S. Rungta Dec 21 '14 at 15:13
  • $\begingroup$ No it wasn´t me :/ I think I dont even have enough rep $\endgroup$ – ViktorG Dec 21 '14 at 15:13
  • $\begingroup$ No, I didn't downvote. $\endgroup$ – David Mitra Dec 21 '14 at 15:14
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We can use approximations : $\ln(x) \approx x-1$ when $x$ is close to $1$. Hence the required limit reduces to $\lim_{x \to 1}(1-x)=0$

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Here are the steps $$ \lim\limits_{x\to 1} \frac{(\ln x)^2}{1-x} = \lim\limits_{x\to 1} \frac{\frac{d}{dx}(\ln x)^2}{\frac{d}{dx}[1-x]} $$ $$ =\lim\limits_{x\to 1} \frac{2(\ln x)\frac{d}{dx}[\ln x]}{\frac{d}{dx}[1]-\frac{d}{dx}[x]} =2\lim\limits_{x\to 1} \frac{\frac{\ln x}{x}}{0-1} $$ $$ =-2\lim\limits_{x\to 1} \frac{\ln x}{x} =-2\left(\frac{0}{1}\right)=-2(0)=0$$

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by L'Hospital we get $$\lim_{x \to 1}2\ln(x)\frac{1}{x}(-1)=0$$

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  • $\begingroup$ yup got it thank you :) $\endgroup$ – ViktorG Dec 21 '14 at 15:00
  • $\begingroup$ It's spelled "L'Hopital." $\endgroup$ – fluffy Dec 21 '14 at 21:24
  • $\begingroup$ i think no, see here mathworld.wolfram.com/LHospitalsRule.html $\endgroup$ – Dr. Sonnhard Graubner Dec 21 '14 at 21:26
  • $\begingroup$ @fluffy It can be either L'Hospital or L'Hôpital. In French, the circumflex indicates that there was a following s in a historical spelling. $\endgroup$ – wchargin Dec 21 '14 at 23:13
  • $\begingroup$ @WChargin Ah, thanks, I didn't know that. I'll drop the pet peeve. :) $\endgroup$ – fluffy Dec 22 '14 at 6:54

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