I need to evaluate $$\lim_{x \to \infty}\ln\left(\frac{1 - x}{1 + x}\right)$$ and I always end up with indetermined form. For example since the following holds true $$\frac{1 - x}{1 + x} = \frac{1 + x - 2x}{1 + x} = 1 - \frac{2x}{1 + x} = 1 - \frac{x}{x} \cdot \frac{2}{\frac{1}{x} + 1} = 1 - \frac{2}{\frac{1}{x} + 1},$$ the limit can be rewritten as
$$\lim_{x \to \infty}\ln\left(\frac{1 - x}{1 + x}\right) = \lim_{x \to \infty}\left(\ln\left(1 - \frac{2}{\frac{1}{x} + 1}\right)\right) = \ln\left(-1\right) = \pi i,$$ but I'm in a regular calculus and can't use $i$'s.
Not really an answer to the question, but rather an answer based on additional info given by @thomashauser in a comment. The "full question" asks to find the asymptotes of
$$ h(x) = \ln\left(\frac{1 - x}{1 + x}\right) \frac{1}{x} $$
First, find the domain: easy to see that the function makes sense only for $|x|<1$. Therefore, you are asked to find the "vertical asymptotes", not the asymptotes at infinity: there is no need to consider the limit $x \rightarrow \infty$.
Clearly you have two asymptotes at $x=-1$ and $x=1$, both going to negative infinity. You may wonder if there is an additional vertical asymptote at $x=0$. The answer is no: in $x=0$ the function is well defined (the limits $x \rightarrow 0^\pm$ coincide). In fact, you can see that close to the origin
$$ h(x) = -2 -2x^2/3+O(x^3) \, .$$
Final note: $h(x)=h(-x)$, i.e. it is an even function, so you just have to study the limits $x \rightarrow 0^+$ and $x \rightarrow 1^-$ (or, equivalently, $x \rightarrow 0^-$ and $x \rightarrow -1^+$).