TITLE: Why does my intuition for "order of divergence" for algebraic fractions fail?
QUESTION [5 upvotes]: I come across this identity once in a while but I actually never grasped it:
$$\frac{2}{1-x^2}=\frac{(1-x)+(1+x)}{(1+x)(1-x)}=\frac{1}{1+x}+\frac{1}{1-x}$$
I'm surprised by it because I would naively not assume that the "order of divergence" at $x=1$ can by reciprocal quadratically and  reciprocal linearly at the same time. I feel one is steeper than the other so the divergence should be quantitatively different. Of course, it is how it is and this just means my intuition for "order of divergence" is an invalid one. I guess I got it from talk like "the integral diverges only logarithmically" and stuff like that.
I wonder if there is a notion of order of divergence for rational function like above. Thoughts: Does partial fraction decomposition say something about which function can't be express to look like they diverge as $\tfrac{1}{1-x}$ on their cricial points? Does it relate to complex analysis= - the residum happens to be concerned with that part of the function too. But my intuition is lacking there too. Is there some resonable $O(x^{-n})$-talk to understand the above identity?

REPLY [3 votes]: Some calculation is involved, but here is one way to view the situation intuitively:
The derivative of $x^2-1$ is non-zero at $x=1$. So near $x=1$, the function $x^2-1$ "behaves like" $a(x-1)$ for some non-zero $a$.