Question: Let's take the function $f(x)=x^{-1}$. My question is this: Is the sum $$\frac{1}{x-a}+\frac{1}{x+a}$$ as $a\rightarrow 0$ near the point $x=0$ equal 0? And how would I proove this anytime I encounter this kind of limit? How would this work with derivatives or integrals?

Answer: 
The quantity you seem to be asking about is $$\lim\_{a\to 0}\frac{1}{x-a}+\frac{1}{x+a}$$ "near" $x=0.$ Well, assuming that $x$ is not $0,$ then this is simply $$\lim\_{a\to 0}\frac{1}{x-a}+\frac{1}{x+a}=\frac1x+\frac1x=\frac2x$$ by continuity. Now, if $x=0,$ then we have $$\lim\_{a\to 0}\frac{1}{x-a}+\frac{1}{x+a}=\lim\_{a\to 0}-\frac{1}{a}+\frac{1}{a}=\lim\_{a\to 0}0=0.$$

This is a good example of why we can't always interchange limits, since $$\lim\_{a\to 0}\lim\_{x\to 0}\frac{1}{x-a}+\frac{1}{x+a}=0,$$ but $$\lim\_{x\to 0}\lim\_{a\to 0}\frac{1}{x-a}+\frac{1}{x+a}$$ does not exist.
Well, it depends what limit are you evaluating first.

First, $\lim\_{a\to 0}(\lim\_{x\to 0} (\frac{1}{x-a}+\frac{1}{x+a}))=\lim\_{a\to 0}0=0$, but, on the other hand, you have $\lim\_{x\to 0}(\lim\_{a\to 0} (\frac{1}{x-a}+\frac{1}{x+a}))=\lim\_{x\to 0}\frac{2}{x}$ and this limit does not exist. So, the order of taking these "iterated" limits in a two variable (or generally, $n$-variable) problems is important.