Find $\lim_\limits{x\to 1^-}{\ln x\cdot \ln(1-x)}$. Find $\lim_\limits{x\to 1^-}{\ln x\cdot \ln(1-x)}$. I can't even start because I don't really know what $x\to 1^-$ means. If you know what it means it would really help me. I would as well if you helped me with finding the limit, hinted or thorough.

$$
\lim_{x\uparrow1} \ln x \ln(1-x) = \lim_{x\uparrow1} \frac{\ln x}{x-1}\cdot\frac{\ln(1-x)}{1/(1-x)}.
$$
Apply L'Hopital's rule to both fractions and you've got it. (In the first one, the numerator and denominator both approach $0$; in the second, each approaches either $+\infty$ or $-\infty$.)
