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How to solve? $\lim_{x\rightarrow 0} \frac{\ln(x)}{1-x}$ I can't use any L'Hôpital or Cauchy rules, only basic limits operations.

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    Can you confirm the limit is$0$or 1 for $x$? If it is $0$ you need no tricks, only the knowledge that $\ln x \to -\infty$ as $x\downarrow 0$.2012-11-07

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Hint: $\lim\limits_{x\to0}\ln(x)=-\infty$

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    Actually I got stuck in that, I don't know why I didn't simply think in that. When things are simple it seems that we complicate them.2012-11-07
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$ \lim_{x \rightarrow 0} \frac{\ln x}{1-x} = \left(\lim_{x \rightarrow 0} \ln x\right)\left( \lim_{x \rightarrow 0} \frac{1}{1-x}\right) = \lim_{x \rightarrow 0} \ln x = -\infty $ If you meant $\lim_{x \rightarrow 1} \ln x /\left(1-x\right)$, $ \lim_{x \rightarrow 1} \frac{\ln x}{1-x} = \lim_{x \rightarrow 1} \frac{\frac{d}{dx}\ln x}{\frac{d}{dx}\left(1-x\right)} = - \lim_{x \rightarrow 1} \frac{1}{x} = -1 $