How to quickly realize that $f(z) \sim \frac{A}{z^k}$, when $z \to \infty$ I have this function:
$$
f(z) = \frac{1}{z+2} \varphi(z)
$$
Where $\varphi(z)$ is a regular branch of $\sqrt{\frac{z}{6-z}}$, such that $\varphi(2+i0) = +\sqrt{\frac{1}{2}}$
What I tried to do was to look how how $f(x)$ behaves as $x \to +\infty$:
$$
f(x) = \frac{1}{x+2}\frac{\sqrt{x}}{i\sqrt{x-6}} \sim \frac{1}{ix} = -\frac{i}{x}
$$
Therefore, I thought $f(z) \sim -\frac{i}{z}$.
Whoever, it turns out I am wrong and $f(z) \sim \frac{i}{z}$.
Where did I make a mistake? How can I quickly calculate the asymptote of such functions at infinity?
 A: Because $z=6$ is a singularity of $\sqrt{\frac{z}{6-z}}$ (where the branch has to be cut), it is unclear what's happening when $z$ crosses $6$ from $6-\epsilon$.
This question doesn't have a definite answer, as it depends on where the branch is cut. Each branch cut is made through a (simple) path from $0$ to $6$. And the cut can be e.g. the upper semi-circle of $|z-3|=3$ or the lower semi-circle of $|z-3|=3$. For these two cuts, the answer will be different.
One method is to to consider $\sqrt{\frac{z}{6-z}}$ as a composition of fraction and square-root, and try to identify which branch of square-root we are really taking. When $z=2$, $\frac{z}{6-z}=\frac{1}{4}$, therefore the square-root branch is exactly the one that extends the positive real root of positive numbers. Depending where the cut is made, this branch can either have $\sqrt{-1}=i$ (cut below the $x$-axis) or $\sqrt{-1}=-i$ (cut above the $x$-axis). If we know which one it is based on the cut, when $z\rightarrow\infty$, we have $\frac{z}{6-z}\rightarrow -1$, hence we know what $\sqrt{\frac{z}{6-z}}\rightarrow \sqrt{-1}$ is.
