How do I prove (without L'Hôpital's rule) that $\lim_{x→0} x⋅\ln x = 0$.
I'm trying to get some intuitive sense for this, but it's quite hard. It's like trying to prove that $x$ goes faster to $\ 0 \ $ then $\ln x$ goes to $-∞$, right ?
I tried this, for $x∈(0,1)$ $x \ln x=x⋅-∫_x^1 \frac{1}{t}dt>x⋅\frac{(1-x)}{-x}=-1+x$
So when $x→0$, I conclude that $x⋅\ln x≥-1$.