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.

here is a way to avoid l'hospitals but will use the fact $\lim_{n \to \infty} \dfrac{\ln n}{n} = 0$  let us do a change of variable $x = 1 - \dfrac{1}{n}.$ so  $$\ln x \ln (1- x) = \ln(1 - 1/n) \ln(1/n)=\ln n(-\ln(1- 1/n)=\ln n\int_{1-1/n}^1\dfrac{dt}{t}$$
we can use the bound $\dfrac{1}{n} \le \int_{1-1/n}^1\dfrac{dt}{t}  \le \dfrac{1}{n-1}$ to conclude $$ \dfrac{\ln n}{n} \le  \ln x \ln (1- x) \le \dfrac{\ln n}{n-1} $$
therefore the $$\lim_{x \to 1_-} \ln x \ln (1- x) = 0$$
