$\lim_{u\rightarrow 0^+}\ln(1-u)\ln u=?$

首先不妨先做个观察, 令 $t=\frac{1}{u}$, $t\rightarrow +\infty$, 从而

\[
\ln(1-u)\ln u=\ln(1-\frac{1}{t})\ln\frac{1}{t}=-\ln(1-\frac{1}{t})\ln t < -\ln(1-\frac{1}{t})\cdot t=\ln(1+\frac{1}{-t})^{-t}
\]

而

\[
\lim_{t\rightarrow +\infty}\ln(1+\frac{1}{-t})^{-t}=1
\]

于是, 当 $t\rightarrow +\infty$ 时,

\[
-\ln(1-\frac{1}{t})\ln t=-\ln(1-\frac{1}{t})\cdot t\cdot \frac{\ln t}{t}\rightarrow 0.
\]

推论: $\lim_{u\rightarrow 1^-}\ln(1-u)\ln u=0$.