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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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    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.2011-06-18

2 Answers 2

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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}$

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    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.2011-06-18
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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.