Problem
$$ { 求极限：\lim_{x\rightarrow 1^-} \ln x\ln \left( 1-x \right) .} $$
Tips Answer
Tips:
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{ 分析：由于这是一个0\cdot \infty 未定式极限，将其变为\frac{\infty}{\frac{1}{0}}=\frac{\infty}{\infty}型极限后利用洛必达法则求解即可.}
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Answer:
$$ { 解：\lim_{x\rightarrow 1^-} \ln x\ln \left( 1-x \right) \begin{array}{l} =\lim_{x\rightarrow 1^-} \frac{\ln \left( 1-x \right)}{\frac{1}{\ln x}}{\mathrm{L}\prime}\lim_{x\rightarrow 1^-} \frac{\frac{1}{1-x}\cdot \left( -1 \right)}{-\frac{1}{\ln ^2x}\cdot \frac{1}{x}}=\lim_{x\rightarrow 1^-} \frac{x\ln ^2x}{1-x}=\lim_{x\rightarrow 1^-} \frac{1\cdot \ln ^2x}{1-x}\\ =\lim_{x\rightarrow 1^-} \frac{\ln ^2x}{1-x}{\mathrm{L}\prime}\lim_{x\rightarrow 1^-} \frac{2\ln x\cdot \frac{1}{x}}{-1}=-\lim_{x\rightarrow 1^-} \frac{2\ln x}{x}=-\frac{2\ln 1}{1}=0.\\ \end{array}} $$
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