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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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What tools can you use? Do you know what are Taylor's polynomials or L'Hopital rule? – Dennis Gulko Nov 7 '12 at 20:45
Did you mean $x\to 0$ or $x\to 1$? – copper.hat Nov 7 '12 at 20:46
The limit isn't even indeterminate. Surely you must've at least tried something before asking this here. What are your thoughts? – EuYu Nov 7 '12 at 20:47
I would suspect the limiting value of $x$ is incorrect and the intent was L'Hopitals... – copper.hat Nov 7 '12 at 20:47
Just updated question. – João Reis Nov 7 '12 at 20:52
up vote 4 down vote accepted

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. – João Reis Nov 7 '12 at 20:56

$$ \lim_{x \rightarrow 0} \frac{\ln x}{1-x} = \left(\lim_{x \rightarrow 0} \ln x\right)\left( \lim_{x \rightarrow 0} \frac{1}{1-x}\right) = \lim_{x \rightarrow 0} \ln x = -\infty $$ If you meant $\lim_{x \rightarrow 1} \ln x /\left(1-x\right)$, $$ \lim_{x \rightarrow 1} \frac{\ln x}{1-x} = \lim_{x \rightarrow 1} \frac{\frac{d}{dx}\ln x}{\frac{d}{dx}\left(1-x\right)} = - \lim_{x \rightarrow 1} \frac{1}{x} = -1 $$

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