Evaluating: $ \lim\limits_{x\to 1^{-}}\left[\ln\left(1-x \over 1 + x\right) - \ln\left(1 - x^{2}\right)\right] $ I would like to evaluate the following limit:
$$
\lim_{x\to 1^{-}}\left[%
\ln\left(1-x \over 1 + x\right) - \ln\left(1 - x^{2}\right)\right]
$$
I put this into a limit calculator and the calculator gave as answer $-\log\left(4\right)$. Why ?.
Should the answer not be?:
$$
\lim_{x \to 1^{-}}\ln\left(1 - x \over 1+x\right)
-\lim_{x \to 1^{-}}\ln\left(1 - x^{2}\right)
=-\infty - \left(-\infty\right) = -\infty + \infty$$
My second, and to me more important question, what is the value of $$-\infty + \infty$$
Thanks for your help.

The limits of the form $\infty - \infty$ are indeterminate. For example


*

*$\lim_{x \to \infty} x - x = 0$,

*$\lim_{x \to \infty} x^2 - x = \infty$,

*$\lim_{x \to \infty} x - x^2 = -\infty$.


In your particular case, observe that


*

*$1 - x^2 = 1^2 - x^2 = (1-x)(1+x)$,

*$\log x - \log y = \log \frac{x}{y}$.


I hope this helps $\ddot\smile$
