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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.

Zev Chonoles
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Nedellyzer
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6 Answers6

13

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 $$ \left|\lim_{x\to 0^+}x^2\ln x\right|\le \lim_{t\to+\infty}\left|\frac{-t}{(e^t)^2}\right|=\lim_{t\to+\infty}\frac{t}{(e^t)^2}\le \lim_{t\to+\infty}\frac{t}{(t+1)^2}=0.$$

Hagen von Eitzen
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10

Sandwich theorem, using the fact (as far as I know it is new)

$$ |\ln(x)| < \frac{1}{x}\,, $$

when $x$ close to $0$, we have

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

Mhenni Benghorbal
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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.

Julien
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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
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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 02 '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. –  Jul 02 '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 02 '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. –  Jul 02 '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 02 '13 at 20:07
  • @MhenniBenghorbal Thanks a lot. –  Jul 03 '13 at 17:42
  • Well done, once again! – amWhy May 27 '14 at 16:44
2

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.

Ovi
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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.

Alex
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