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.

To prove that 
\begin{equation*}
\lim_{x\rightarrow 1^{-}}\ln x\ln (1-x)=0,
\end{equation*}
it is possible to avoid, L'Hospital's rule, Taylor series and also to avoid
the limits
\begin{equation*}
\lim_{u\rightarrow 0}u\ln u=0,\ \ \ \ \text{and}\ \ \ \ \ \ \ \
\lim_{n\rightarrow +\infty }\frac{\ln n}{n}=0.
\end{equation*}
It suffices to use the well-known inequality
\begin{equation*}
0\leq \ln x\ln (1-x)\leq \sqrt{x(1-x)},\ \ x\in (0,1).
\end{equation*}
and the squeeze theorem.
PS. An elegant proof of the well known inequality cited above maybe found here: 
with this inequality $\ln{x}\ln{(1-x)}<\sqrt{x(1-x)}$.
