Finding $\lim\limits_{x\to0}x^2\ln (x)$ without L'Hospital

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

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

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

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.