# 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 $\mathbb{R} \setminus \{ -2, 2\}$.

• I'm not entirely sure what you're asking. If you want to understand how $\varepsilon$-$\delta$-proofs work, you could do worse than looking at explicit examples. You can find quite a few of these by searching on the site, e.g. using this search. – t.b. Dec 5 '11 at 13:33
• And maybe this is what you're looking for? – t.b. Dec 5 '11 at 13:35
• You should study this answer, found in the search that t. b. suggests: math.stackexchange.com/a/11884/7850 – The Chaz 2.0 Dec 5 '11 at 13:45
• eps-delta is traditionally hard because it is one of the first times you encounter a statement with nested exists and for-all quantifiers. The problem is that unless you can specify a specific problem you are having I don~t think anyone here will come up with an explanation that is much better then the one in your textbook. – hugomg Dec 5 '11 at 13:49
• The $\epsilon$-$\delta$ is more unpleasant than usual in this case. In outline it is not hard, but there are petty details to worry about "near" $\pm 2$. Showing continuity at (say) $x=1.9$, or some other concrete number, would enable you to concentrate on the basic idea of the proof. – André Nicolas Dec 5 '11 at 18:31

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.
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.