15
$\begingroup$

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)

$\endgroup$
  • 1
    $\begingroup$ Observing that $x\ln x=\ln(x^x),$ this question is effectively a duplicate of this other one. $\endgroup$ – Cameron Buie Oct 11 '13 at 21:57
  • 8
    $\begingroup$ I do not believe this should be closed, since it describes an interesting aproach to the problem that is absent elsewhere. $\endgroup$ – André Nicolas Oct 11 '13 at 22:10
  • 3
    $\begingroup$ Lovely boldfaced typo. Have a sticky p key. Also shift. $\endgroup$ – André Nicolas Oct 11 '13 at 22:22
  • $\begingroup$ @Raekye : That is a very clever approach. $\endgroup$ – Stefan Smith Oct 12 '13 at 1:54
  • $\begingroup$ @AndréNicolas: Couldn't the approach be posted to the other post? I think it would make sense. $\endgroup$ – Najib Idrissi Oct 12 '13 at 2:55
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – Raekye Oct 11 '13 at 22:12
  • $\begingroup$ 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 $0<x<1$. $\endgroup$ – Jonas Meyer Oct 11 '13 at 22:15
  • $\begingroup$ (Now that $f(x)=x\ln(x)$ instead of $2x\ln(x)$, the last line of my comment should say "$|f(x)|\leq 1$".) $\endgroup$ – Jonas Meyer Oct 11 '13 at 22:57
  • $\begingroup$ Sorry about the little change, in checking for my usual typos I thought there was no point in dragging the $2$ around. $\endgroup$ – André Nicolas Oct 11 '13 at 22:59
  • $\begingroup$ @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. $\endgroup$ – André Nicolas Oct 12 '13 at 0:27
9
$\begingroup$

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 $L=0$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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