Originally Posted by

**Mathstud28** I have to do it...I dont know advanced you are, but if you have gotten to infinite series

$\displaystyle \ln(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}, \forall{x}\in[0,2)$

and since $\displaystyle 0^{+}$ is contained in this

$\displaystyle \therefore\sqrt{x}\ln(x)=\sqrt{x}\cdot\sum_{n=0}^{ \infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}$

as you can see every term will have an x in it, thus direct substitution will yield $\displaystyle 0+0+0+...+0=0$

$\displaystyle \therefore\lim_{x\to{0^{+}}}\sqrt{x}\ln(x)=0$

EDIT: Also, are you aware of the conept of higher order infinitesimals with limits, $\displaystyle \sqrt{x}\succ\ln(x)$?