$\lim_{x\to 0+} (x\ln(x))$ without L'Hôpital's rule

Question

Is there a way to find $\lim_{x\to 0+} (x\ln(x))$ without using L'Hôpital's rule? I'm trying to solve this with squeeze theorem but couldn't figure out a lower bound.

Answer 1

Assuming that $\ln (1+t) \leq t$ for $t >0$ you can prove this using the following inequalities: $0\leq -x \ln x =2x\ln (\frac 1 {\sqrt x}) \leq 2x( \frac 1 {\sqrt x}-1) =2\sqrt x -2x$ for $0 <x<1$.