How do i compute the following 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)}} $
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}$
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.