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

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
$${-{1 \over 4}}\ln(1) + {1 \over 2}\ln(1) +{\pi i \over 2} $$
$$ = {\pi i\over 2}$$
