Help with epsilon-delta proof of continuity I just asked a specific homework question on this topic, but I want a more general explanation for how to go about proving continuity with this method.
I can't even wrap my head around what the proof is really saying, let alone figure out the steps to prove a specific function is continuous.
Edit: okay, here is a specific example. Maybe a walkthrough would be easier for you guys to help me out:

Prove that $g(x) = \frac{1}{x^2 - 4}$ is continuous on $\mathbb{R} \setminus \{ -2, 2\}$. 

 A: So, consider $g(x) = 1/(x^2-4)$, and let's show it is continouous at $1.9$.  
Let $\epsilon > 0$ be given.  Define $\delta = $ (...to be filled in later...)  
Now, let $x$ be such that $|x-1.9|<\delta$.  Then ... (details to be filled in) ... So we have $|g(x) - g(1.9)| < \epsilon$. This shows $g$ is continuous at $1.9$, QED.  
What can you fill in next to improve this and make it closer to the final thing we want?  [Note: Community Wiki, so others may edit.]
A: To avoid the problems at $2$ and $-2$ let's consider what happens with $\lim_{x \to 1}10x$.  Intuitively, when you say this is $10$, we agree that as $x$ gets close to $1, 10x$ gets close to $10$.  The $\epsilon-\delta$ formulation can be thought as a game.  If you claim it is $10$, you say that whatever $\epsilon \gt 0$ I name, you can say how close $x$ has to be to $1$ so that $10x$ is within $(10-\epsilon,10+\epsilon)$.  For this simple example, it is easy to see.  You say that $\delta=\epsilon/11$ (or something convenient smaller than $\epsilon/10$).  If you had tried to claim the limit was $9.999$ I could win by taking $\epsilon=0.0001$.  Proving it over a range, like $\mathbb{R} \setminus \{2,-2\}$ in your example is the same idea, but it has to work with all the $x$ in the range, but the $\delta$ can depend upon $x$ as well as $\epsilon$.  As you get close to $2$ the slope of the curve gets very high, so $\delta$ will have to be very small.
