Computing the limit $\lim_{x\to\infty}{-\frac{1}{4}\ln{(1+x^2)}+\frac{1}{2}\ln{(1-x)}}$

How do i compute the following limit?

$$\lim_{x\to\infty}{-\frac{1}{4}\ln{(1+x^2)}+\frac{1}{2}\ln{(1-x)}}$$

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Your question is unclear, because I think you are missing parens. In the absence of these, your answer would b $+\infty$ since the $\ln(1)$ would evaluate to zero. –  ncmathsadist Jun 18 '11 at 15:13
Something doesn't seem right—$\ln(1-x)$ is only defined (in a real-analysis sense) for $x<1$... –  Isaac Jun 18 '11 at 15:18
Perhaps that $2$ outside was a mistken simplification, and it was really $-(1/4)\ln(1+x^2) + \ln((1-x)^2)$ But we won't know unless one-question Julius returns. –  GEdgar Jun 18 '11 at 16:17
Yeah, sorry about that. I have fixed the tags and the function. Now it has a finite, complex value (namely i*Pi/2), the question still stands. –  Julius Jun 18 '11 at 16:28

Taking $\log(-a) = \pi i + \log(a)$ for $a > 0$, the limit becomes $$\lim_{x \rightarrow \infty}\bigg({-{1 \over 4}}\ln(1 + x^2) + {1\over 2}\ln(x - 1) + {\pi i\over 2}\bigg)$$ Note that $\ln(1 + x^2) = \ln(({1 \over x^2} + 1)(x^2)) = \ln({1 \over x^2} + 1) + 2\ln(x)$, and that $\ln(x - 1) = \ln((1 - {1 \over x} )(x)) = \ln(1 - {1 \over x}) + \ln(x)$. So the limit is the same as $$\lim_{x \rightarrow \infty}\bigg({-{1 \over 4}}\ln({1 \over x^2} + 1) + {1 \over 2}\ln(1 - {1 \over x}) + {\pi i \over 2}\bigg)$$ The functions here converge to finite limits as $x$ goes to infinity. You get $${-{1 \over 4}}\ln(1) + {1 \over 2}\ln(1) +{\pi i \over 2}$$ $$= {\pi i\over 2}$$
I'm taking the logarithm of $-1$ to be $\pi i$ here, which agrees with the usual branches. Otherwise the answer can be adjusted accordingly –  Zarrax Jun 18 '11 at 17:01
I know that the logarithm is defined on $\Bbb{C}\setminus (-\infty,0]$, and therefore, you cand take logarithms of negative numbers. I think that the OP has put the wrong signs in the second term. There's no way someone which takes complex analysis courses and therefore has passed some classic analysis course wouldn't be able to solve this limit. –  Beni Bogosel Jun 18 '11 at 18:53
Now you have added parens. This has no limit since, when $x\to\infty$, $1 - x$ becomes negative and is outside the domain of the log function.