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.
 A: $$\frac{\log(1-x)}{\frac1{\log x}}\stackrel{l'H}\rightarrow\frac{-\frac1{1-x}}{-\frac{\frac1x}{\log^2x}}=\frac{x\log^2 x}{1-x}\stackrel{l'H}\rightarrow\frac{\log^2x+2\log x}{-1}\xrightarrow[x\to1^-]{}0 $$
A: With equivalents, it's very short: set $x=1-u\enspace(u>0)$. Then $\ln x\ln(1-x)=\ln(1-u)\ln u$.
As $\ln(1-u)\sim_0 -u$, we have $\,\,\ln(1-u)\ln u\sim_0-u\ln u$, which tends to $0$ as $u$ tends to $0$.
Hence $$\lim_{x\to 1_-}\ln x\,\ln(1-x)=\lim_{u\to 0_+}\ln(1-u)\ln u=0.$$
A: $$
\lim_{x\uparrow1} \ln x \ln(1-x) = \lim_{x\uparrow1} \frac{\ln x}{x-1}\cdot\frac{\ln(1-x)}{1/(1-x)}.
$$
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 $-\infty$.)
A: 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 $$\lim_{x \to 1_-} \ln x \ln (1- x) = 0$$
A: 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 $\ln{x}\ln{(1-x)}<\sqrt{x(1-x)}$.
