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

  • 1
    $\begingroup$ 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. $\endgroup$ – t.b. Dec 5 '11 at 13:33
  • 1
    $\begingroup$ And maybe this is what you're looking for? $\endgroup$ – t.b. Dec 5 '11 at 13:35
  • 3
    $\begingroup$ You should study this answer, found in the search that t. b. suggests: math.stackexchange.com/a/11884/7850 $\endgroup$ – The Chaz 2.0 Dec 5 '11 at 13:45
  • $\begingroup$ 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. $\endgroup$ – hugomg Dec 5 '11 at 13:49
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.

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.]


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.