I'm blanking out, how would I show either a sequential limit or epsilon-delta proof for this? For the sequential limit definition, is it an obvious property that if , then . If not, how do I go about proving that, or how would I show an epsilon-delta proof? I am completely stuck on how it could be shown using epsilon-delta. Thanks.
I'm sorry, I don't see how that helps. Are you trying to use that for an epsilon delta proof that is continuous on ? Because then I want to show that there exists delta such that for
Err, nevermind, I caught your error and with the correction I understand it.
Okay, so I worked out if , then with manipulation we get if then
I think that's correct, right?
Theorem: Let be such that is continuous at and then is continuous on .
If we're done since and is trivially continuous. Thus, assume that then . Also, notice that for any since we covered the case above assume that then which is obviously a contradiction. It follows that and so
Thus, since is continuous at for every there exists some such that .
So, let be given and arbitrary. Then,
Now, let . We know that there exists a such that . Thus, if . It follows that is continuous at .
Since was arbitrary it follows that is continuous on . .
All you have to do is prove that is continuous at
It might be interesting to point out that while I showed the proof I did because it is interesting it turns out that the only continuous function such that (and thus the only function of this form which is continuous at as I prove above) is .
To see this first note that by induction we have that for every that . Therefore, if we have that . THus, we have that if that .
It follows that if we define we have that is dense in and since it's closed it follows that from where the conclusion follows.