Find $\lim_{x\to 0^+} \ln x\cdot \ln(1-x)$ Find $$\lim_{x\to 0^+} \ln x\cdot \ln(1-x)$$
I've been unable to use the arithmetic rules for infinite limits, as $\ln x$ approaches $-\infty$ as $x\to 0^+$, while $\ln(1-x)$approaches $0$ as $x\to 0^+$, and the arithmetic rules for the multiplication of infinite limits only applies when one of the limits is finite and nonzero.
Can anyone point me in the right direction for finding this limit? I've been unable to continue..
(Spoiler: I've checked WolframAlpha and the limit is equal to $0$, but this information hasn't helped me to proceed)

Hint: Another approach which is similar to @Mhenni's is:
When $x\to 0$ and we know that $\alpha(x)$ is very small then $$\ln(1+\alpha(x))\sim\alpha(x)$$
