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?
The way I see it is very simple: your left hand term is $$ \frac2{1-x^2}=\frac2{(1-x)(1+x)}=\frac2{1+x}\,\frac1{1-x}. $$ The first factor on the right is not blowing up. All the divergence is coming from the factor $1/1-x$.
The same happens in your right hand term: the first summand converges to $1/2$ and so it is not contributing to the divergence.
Edit: Something that might help avoid confusion is to move the limit to the origin. If we set $t=x-1$, then $1-x^2=1-(t+1)^2=-t^2-2t$. So we are considering the expression $$ \frac2{-t^2-2t}=-\frac2{2t+t^2}. $$ When $t$ is very near $0$ (which is the case when $x$ is very near $1$) this expression is roughly $-1/t$.