Evaluate $\lim\limits_{x\to 1} \frac{(\ln x)^2}{1-x}$

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?

• No it doesn't. It means just that: the limit is $\frac{0}{-1}=0$ – SBareS Dec 21 '14 at 14:55
• So it just means that it converges to $0$ ? thats it ? – ViktorG Dec 21 '14 at 14:56
• Yep, that's it. – SBareS Dec 21 '14 at 14:56
• It's spelled "L'Hopital." – fluffy Dec 21 '14 at 21:27
• @fluffy my bad, changed it – ViktorG Dec 21 '14 at 21:29

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.

• 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 ... – ViktorG Dec 21 '14 at 15:08
• @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. :) – Ahaan S. Rungta Dec 21 '14 at 15:12
• @ViktorG Did you downvote? I suggest you revert now. :) – Ahaan S. Rungta Dec 21 '14 at 15:13
• No it wasn´t me :/ I think I dont even have enough rep – ViktorG Dec 21 '14 at 15:13
• No, I didn't downvote. – David Mitra Dec 21 '14 at 15:14

We can use approximations : $\ln(x) \approx x-1$ when $x$ is close to $1$. Hence the required limit reduces to $\lim_{x \to 1}(1-x)=0$

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

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

• yup got it thank you :) – ViktorG Dec 21 '14 at 15:00
• It's spelled "L'Hopital." – fluffy Dec 21 '14 at 21:24
• i think no, see here mathworld.wolfram.com/LHospitalsRule.html – Dr. Sonnhard Graubner Dec 21 '14 at 21:26
• @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. – wchargin Dec 21 '14 at 23:13
• @WChargin Ah, thanks, I didn't know that. I'll drop the pet peeve. :) – fluffy Dec 22 '14 at 6:54