# Solving $\lim_{x\rightarrow 0} \frac{\ln(x)}{1-x}$

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.

-
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
show 1 more comment

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

-
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 $\lim_{x \rightarrow 1} \ln x /\left(1-x\right)$, $$\lim_{x \rightarrow 1} \frac{\ln x}{1-x} = \lim_{x \rightarrow 1} \frac{\frac{d}{dx}\ln x}{\frac{d}{dx}\left(1-x\right)} = - \lim_{x \rightarrow 1} \frac{1}{x} = -1$$
I'm going to assume that you meant $x\to 1$, rather than $x\to 0$ (since otherwise, it's relatively easy). In that case, we have an indeterminate limit of the form "$\frac00$". If you know L'Hopital's rule, this would be an ideal place to use it. If you'd rather not do that, and you know about Taylor polynomials, then find the Taylor polynomial of $\ln x$ about $x=1$, note that $1-x$ can be factored out of it, then evaluate the remaining series at $x=1$.