Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
6  
Did you mean $x\to 0$ or $x\to 1$? –  copper.hat Nov 7 '12 at 20:46
1  
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
1  
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

2 Answers 2

up vote 4 down vote accepted

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

share|improve this answer
    
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 $\lim_{x \rightarrow 1} \ln x /\left(1-x\right)$, $$ \lim_{x \rightarrow 1} \frac{\ln x}{1-x} = \lim_{x \rightarrow 1} \frac{\frac{d}{dx}\ln x}{\frac{d}{dx}\left(1-x\right)} = - \lim_{x \rightarrow 1} \frac{1}{x} = -1 $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.