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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.

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it should be $x\to 0$, sorry –  Sjoerd Smaal Jul 2 '13 at 18:36
2  
$\lim_{y \to \infty} -\log y/y^2$. $\log y \leqslant y$ 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 $$ \left|\lim_{x\to 0^+}x^2\ln x\right|\le \lim_{t\to+\infty}\left|\frac{-t}{(e^t)^2}\right|=\lim_{t\to+\infty}\frac{t}{(e^t)^2}\le \lim_{t\to+\infty}\frac{t}{(t+1)^2}=0.$$

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

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

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2  
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
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@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.

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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.

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Using the MVT is the same as using l'Hôpital. –  egreg Jul 2 '13 at 19:35
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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
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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.

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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.

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