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.
 A: The problem shows that, in fact, $-\infty +\infty$ can be anything. You can use properties of logarithms to take this limit.
$\lim_{x\rightarrow 1^{-}}(\ln \frac{1-x}{1+x}-\ln (1-x^2))$
$=\lim_{x\rightarrow 1^{-}}(\ln (1-x) -\ln (1+x)-\ln (1-x)(1+x))$
$=\lim_{x\rightarrow 1^{-}}(\ln (1-x) -\ln (1+x)-\ln (1-x)-\ln(1+x))$
$=\lim_{x\rightarrow 1^{-}}-2\ln (1+x)=-2\ln 2=-\ln 2^2=-\ln 4$.
A: There's a theorem that if $\lim\limits_{x \to a} f(x) = L$ and $\lim\limits_{x \to a} g(x) = M$, then
$$
\lim_{x \to a} \bigl[f(x) + g(x)\bigr] = L + M.
$$
In the equation after "Should the answer not be?", you've applied the theorem as $x \to 1^{-}$ to the functions
$$
f(x) = \ln \left(\frac{1 - x}{1 + x}\right),\qquad
g(x) = -\ln(1 - x^{2}),
$$
for which the hypothesis (existence of both limits) is false. Consequently you have no reason to expect the conclusion of the theorem to hold (or even to be meaningful).
As multiple respondents have noted, a formal expression $\infty - \infty$ is indeterminate. Your limit exists, but it must be evaluated by applying properties of the logarithm, not by using "limit of the sum is the sum of the limits".
A: Your expression $-\infty + \infty$ is an indeterminate form, much like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. So, on its own, it has no meaning.
In your specific case, using the property of logs that $\ln(AB) = \ln(A) + \ln(B)$ will likely be helpful here.
A: Hint:
Remember that $\ln(A) - \ln(B) = \ln(A/B)$.  Further, $\ln(x^a) = a\ln(x)$.  Also, $1-x^2 = (1+x)(1-x)$ is helpful.
Can you solve it from here?
A: 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 $\ddot\smile$
A: Since $1-x^2 = (1-x)(1+x)$ we have, 
$$\lim_{x\to 1^{-}}\left[
\ln\left(1-x \over 1 + x\right) - \ln\left(1 - x^{2}\right)\right] =\left[\ln\left(1-x \right)-\ln\left( 1 + x\right)\right] - \left[\ln\left(1 +x\right)+\ln\left(1 - x\right)\right] = -2\ln\left(1 +x\right)$$
Thus, $$\lim_{x\to 1^{-}}\left[
\ln\left(1-x \over 1 + x\right) - \ln\left(1 - x^{2}\right)\right] = -2\lim_{x\to 1^{-}}\ln\left(1 +x\right) = -2\ln 2= -\ln 4.$$
