Finding $\lim\limits_{x\to0}x^2\ln (x)$ without L'Hospital

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.

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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

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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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. - 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. - 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. – user63181 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. – user63181 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. - 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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