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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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    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.2011-06-18
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    Something doesn't seem right—$\ln(1-x)$ is only defined (in a real-analysis sense) for $x<1$...2011-06-18
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    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.2011-06-18
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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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    Beware of the branch-of-the-log issue.2011-06-18
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    I'm taking the logarithm of $-1$ to be $\pi i$ here, which agrees with the usual branches. Otherwise the answer can be adjusted accordingly2011-06-18
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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.