Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am preparing a resit for calculus and I encountered a limit problem.

The problem is the following: $\lim\limits_{x\to0}x^2\ln (x)$

I am not allowed to use L'Hospital.

Please help me, I am stuck for almost an hour now.

share|improve this question
it should be $x\to 0$, sorry –  Sjoerd Smaal Jul 2 '13 at 18:36
$\lim_{y \to \infty} -\log y/y^2$. $\log y \leqslant y$ for $y \geqslant 1$ is easy to see without l'Hopital. –  Daniel Fischer Jul 2 '13 at 18:39
Since you are not allowed to use L'Hospital, can you tell us what you can use? –  M Turgeon Jul 2 '13 at 18:39
Write it down in a different way, like Daniel Fischer just showed, thank you Daniel –  Sjoerd Smaal Jul 2 '13 at 18:40

6 Answers 6

up vote 12 down vote accepted

Note that we should rather consider $$\lim_{x\to 0^+}x^2\ln x.$$ Substitute $x$ with $e^{-t}$ (this is possible for $x>0$) to get $$ \lim_{x\to 0^+}x^2\ln x=\lim_{t\to+\infty}(-t)(e^{-t})^2=\lim_{t\to+\infty}\frac{-t}{(e^t)^2}$$ and use our favorite estimate for the exponential function: $e^t\ge 1+t$, to get $$ \left|\lim_{x\to 0^+}x^2\ln x\right|\le \lim_{t\to+\infty}\left|\frac{-t}{(e^t)^2}\right|=\lim_{t\to+\infty}\frac{t}{(e^t)^2}\le \lim_{t\to+\infty}\frac{t}{(t+1)^2}=0.$$

share|improve this answer

Sandwich theorem, using the fact $ |\ln(x)| < \frac{1}{x} $, when $x$ close to $0$, we have

$$ |x^2\ln (x)| < x. $$

share|improve this answer
this seems better to fit OP's request. –  chenbai Jul 3 '13 at 4:02
@chenbai: Thanks for the comment. –  Mhenni Benghorbal Jul 3 '13 at 16:54
@Mhenni sir, I am having trouble to understand the line "..using the fact $ |\ln(x)| < \frac{1}{x} \,\,$". I can not prove the result $ |\ln(x)| < \frac{1}{x} \,\,$ Can you explain it ,please? –  learner Aug 27 '13 at 4:51

Just to use FTC: for $0<x\leq 1$$$ 0<-x\ln x=x\int_x^1\frac{1}{t}dt=\int_x^1\frac{x}{t}dt\leq \int_x^11dt=1-x\leq 1\Rightarrow 0<-x^2\ln x\leq x. $$ Hmm... that's Mhenni Benghorbal's argument. But since I prove the inequality, I guess I'll leave it.

share|improve this answer
what is the full form of FTC? Also ,I have little trouble to understand the line $$...=\int_x^1\frac{x}{t}dt\leq \int_x^1 1dt=..$$ .Since $0 <x \leq 1$, $$...=\int_x^1\frac{x}{t}dt\leq \int_x^1 \frac 1 tdt=..$$ is understandable. But I could not understand why or how $t$ is missing. Can you clarify ,please? –  learner Aug 27 '13 at 11:27

By mean value theorem there's $\xi_x\in (x,1)$ such that $$\log(x)-\log(1)=(x-1)\frac{1}{\xi_x}$$ and since $$x\leq\frac{x}{\xi_x}\leq1$$ we have $$x(x-1)\leq x^2 \log x = x^2 (x-1)\frac{1}{\xi_x}\leq x^2(x-1)$$ so we conclude by squeeze theorem.

share|improve this answer
Using the MVT is the same as using l'Hôpital. –  egreg Jul 2 '13 at 19:35
You're right if the MVT is a consequence of the l'Hôpital theorem but this not the case. –  Sami Ben Romdhane Jul 2 '13 at 19:53
Well, one can easily prove l'Hôpital's theorem with the MVT. So you're saying that the OP can prove the theorem and then use it. ;-) –  egreg Jul 2 '13 at 19:59
Ok but I don't proved the l'Hôpital theorem and I just used the MVT and if you don't accept this then you shoudn't accept the Hagen von Eitzen's answer since his crucial inequality $e^t\geq 1+t$ can be proved by the MVT. –  Sami Ben Romdhane Jul 2 '13 at 20:05
I was joking, of course. I can't understand why the poor students are always requested not to use the theorem. It can be fun (but surely tiring) to climb the Tour Eiffel by the stairs; using the lift is much more efficient. –  egreg Jul 2 '13 at 20:07

set $\log x=t$, so the limit is $$ \lim_{t \to \infty}e^{2t}t <\lim_{t \to -\infty}e^{2t +t}=0 $$ Clearly $te^{2t}<0$ for $t<0$. At the same time if you take the derivative of $te^{2t}$ you get $e^{2t}(2t^2+1)$, which is always non-negative, hence the limit of the original function is 0.

share|improve this answer

For an elementary way consider this:

$\lim_{x\to 0} x^2 \ln x$

$\lim_{x\to 0} \ln {e^{x^2}} \ln x$

$\lim_{x\to 0} \ln [(e^{x^2})^{\ln x}]$

$\lim_{x\to 0} \ln[(e^{\ln x})^{x^2}]$

$\lim_{x\to 0} \ln x^{x^2}$

$\lim_{x\to 0} \ln (x^x)^x$

It is a well-known fact that $\lim_{x\to 0} x^x=1$ , so if we substitute that in we get $\lim_{x\to 0} \ln 1^x$ , or $\lim _{x\to 0} \ln 1=0$ . So our original limit is equal to $0$ . Let me know if anything is not clear. Hope this helps.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.