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

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 $L=0$.

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 $\displaystyle\lim_{x\to 0} x^2\ln(x^2)=0$, and therefore $\displaystyle\lim_{t\to 0^+}t\ln t=0$.

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.