I'm having trouble figuring this one out:

$\displaystyle \lim_{x\to0^+}sin(x)ln(x) $

$\displaystyle = \lim_{x\to0^+}\frac{sin(x)}{\frac{1}{ln(x)}} $

$\displaystyle = \lim_{x\to0^+}\frac{cos(x)}{\frac{-1}{x(ln(x))^2}} $

Which doesn't seem to help very much.