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.
Solving $\lim_{x\rightarrow 0} \frac{\ln(x)}{1-x}$
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calculus
limits
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0What tools can you use? Do you know what are Taylor's polynomials or L'Hopital rule? – 2012-11-07
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6Did you mean $x\to 0$ or $x\to 1$? – 2012-11-07
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1The limit isn't even indeterminate. Surely you must've at least tried something before asking this here. What are your thoughts? – 2012-11-07
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1I would suspect the limiting value of $x$ is incorrect and the intent was L'Hopitals... – 2012-11-07
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0Just updated question. – 2012-11-07
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0Can 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
2 Answers
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Hint: $\lim\limits_{x\to0}\ln(x)=-\infty$
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1Actually 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 $\lim_{x \rightarrow 1} \ln x /\left(1-x\right)$, $$ \lim_{x \rightarrow 1} \frac{\ln x}{1-x} = \lim_{x \rightarrow 1} \frac{\frac{d}{dx}\ln x}{\frac{d}{dx}\left(1-x\right)} = - \lim_{x \rightarrow 1} \frac{1}{x} = -1 $$