$\lim_{x\to0^{+}} x \ln x$ without l'Hopital's rule

I have a midterm coming up and on the past exams the hard question(s) usually involve some form of $\lim_{x\to0^{+}} x \ln x$. However, we're not allowed to use l'Hopital's rule, on this year's exam anyways.

So how can I evaluate said limit without l'Hopital's rule? I got somewhere with another approach, don't know if it's useful:

1. $\lim_{x\to0^{+}} x \ln x = \lim_{x\to0^{+}} x^2 \ln (x^2) = L$
2. $= (\lim_{x\to0^{+}} 2x)(\lim_{x\to0^{+}} x \ln x)$
3. $= 0 * L$

Then I just need to prove that L is finite/exists (which means it must be 0)

• Observing that $x\ln x=\ln(x^x),$ this question is effectively a duplicate of this other one. – Cameron Buie Oct 11 '13 at 21:57
• I do not believe this should be closed, since it describes an interesting aproach to the problem that is absent elsewhere. – André Nicolas Oct 11 '13 at 22:10
• Lovely boldfaced typo. Have a sticky p key. Also shift. – André Nicolas Oct 11 '13 at 22:22
• @Raekye : That is a very clever approach. – Stefan Smith Oct 12 '13 at 1:54
• @AndréNicolas: Couldn't the approach be posted to the other post? I think it would make sense. – Najib Idrissi Oct 12 '13 at 2:55

The idea you described is a very nice one. We fill in the details.

We consider, as in the OP, $x^2\ln(x^2)$, that is, $(2x)(x\ln x)$. If we can show that $x\ln x$ is bounded near $0$, it will follow by Squeezing that $\lim_{x\to 0} x^2\ln(x^2)=0$, and therefore $\lim_{t\to 0^+}t\ln t=0$.

Let $f(x)=x\ln x$. Then $f'(x)=1+\ln x$. It follows that $f(x)$ is decreasing in the interval $(0,e^{-1})$. It reaches a minimum value of $-e^{-1}$ at $x=e^{-1}$.

Since $f(x)$ is negative in our interval, we have $|x\ln x|\le e^{-1}$ in the interval, and we have shown boundedness.

• Ah, that makes sense. I thought I had to use the derivative to show "which direction the function is going" but couldn't spell it out. Thank you very much! – Raekye Oct 11 '13 at 22:12
• A deleted answer gives another way to show that $f(x)$ is bounded: $$0>x\ln(x)=x\int_1^x\frac1t\,dt\geq x(\frac1x(x-1))=x-1>-1$$ when $0<x<1$, so $|f(x)|\leq 2$ when $0<x<1$. – Jonas Meyer Oct 11 '13 at 22:15
• (Now that $f(x)=x\ln(x)$ instead of $2x\ln(x)$, the last line of my comment should say "$|f(x)|\leq 1$".) – Jonas Meyer Oct 11 '13 at 22:57
• Sorry about the little change, in checking for my usual typos I thought there was no point in dragging the $2$ around. – André Nicolas Oct 11 '13 at 22:59
• @Raekye: Please note that in my opinion the approach of user@17762 is "better." My answer was an exercise in pushing through your clever idea. – André Nicolas Oct 12 '13 at 0:27

Let $x=e^{-t}$ and note that as $x \to 0^+$, we have $t \to \infty$. Hence, $$L = \lim_{x \to 0} x \ln(x) = \lim_{t \to \infty} -te^{-t} = -\lim_{t \to \infty} \dfrac{t}{e^t}$$ Now recall that $e^t \geq \dfrac{t^2}2$, because $$e^t =\sum_{k=0}^{\infty}\frac{t^k}{k!} \geq \frac{t^2}{2}$$ Hence, we have $$\lim_{t \to \infty} \dfrac{t}{e^t} \leq \lim_{t \to \infty} \dfrac2t = 0$$ This gives us $L=0$.

• Ah this makes sense too. I selected Andre's answer though because he answered earlier. Thanks for your input though! – Raekye Oct 11 '13 at 22:17
• Can you explain why e^t >= t^2/2 ? I’m not good at math. – plhn Mar 15 '17 at 12:10