TITLE: $\lim_{x\to0^{+}} x \ln x$ without l'Hopital's rule
QUESTION [20 upvotes]: 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:

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

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

REPLY [13 votes]: 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 $L=0$.