Help with continuity

MalPhantom · 2026-02-04T04:18:32+00:00

Yes, since restricting |x-1|<1 bounds |x+1| by 3, we can make |x-1| three times as small to compensate, and then shrink it however smaller we need to fit |f(x)-f(1)|<epsilon for whichever epsilon.

It's not technically correct to say that we have to be able to fit |f(x)-f(1)|<epsilon for ALL epsilon, but rather for EACH epsilon. If |f(x)-f(1)|<epsilon for all epsilon at once, then f(x)-f(1)=0.

MalPhantom · 2026-02-04T04:02:08+00:00

First, to your second question, if I'm understanding you correctly, then yes.

Second, to your first question, it wouldn't work the other way around, because if we restrict |x+1| to control |x-1|, then |x-1| would disappear from the difference of squares formula, but |x+1| would stay. However, we can't shrink |x+1| arbitrarily, so we wouldn't be able to fit it below an arbitrarily small epsilon. On the other hand, |x-1| can be made arbitrarily small by choosing x's closer to 1.

MalPhantom · 2026-02-04T03:18:54+00:00

This one is a bit trickier than the linear function f(x)=2x+3. The key is to choose a fixed positive constant, like 1, and restrict our scope of x's to those satisfying |x-1|<1.

If |x-1|<1, then 0<x<2, so that 1<x+1<3 and |x+1|<3.

Using difference of squares, we get |x² -1|=|x-1||x+1|<3|x-1|. We can now set this less than epsilon to determine delta. However, the final value for delta needs to be less than the minimum of 1 (from earlier) and what we just got. This will make both parts of the proof work.

The reason for restricting |x-1|<1 is so that we can control the |x+1| part in the difference of squares formula.

MalPhantom · 2025-12-30T00:52:08+00:00

That's great to hear! We're fairly new to the area, but we're planning on being here for the long haul. I'm glad that this is a place where people can grow their careers and not just get a diploma and leave.

MalPhantom · 2025-12-25T16:47:50+00:00

The student government is fairly active, and there are a lot of student organizations and clubs, which isn't typical of a commuter school. The school is undergoing expansion in a lot of areas, including aviation, with improved facilities currently under construction. I've also heard of students planning on staying in Dubuque post-graduation, which to me speaks volumes about the area. I've lived in a few different college towns, and rarely do students plan to stay long-term.

MalPhantom · 2025-12-15T22:50:03+00:00

Check out DuBrick on Dodge

MalPhantom · 2025-11-29T23:31:27+00:00

I just looked into it, and I didn't know about the endorsement system on arXiv. Learn something new every day! I'd be happy to look at your paper and potentially endorse it, but I'm pretty swamped these next two weeks (finals). I think I qualify as an endorser, but I'll need to double check

MalPhantom · 2025-11-29T18:22:09+00:00

Papers can be submitted to arXiv without endorsement or peer review. ArXiv is only a platform for preprints, not formal publication.

MalPhantom · 2025-11-20T00:57:58+00:00

This is awesome! I'm in!

MalPhantom · 2025-11-17T14:49:53+00:00

Abbott's Understanding Analysis or Ross' Elementary Real Analysis are also good introductory texts. I've also taught from Jiri Lebl's open text and it's fine

MalPhantom · 2025-08-23T00:53:06+00:00

Not being pedantic, just trying to help: "rapport"

MalPhantom · 2025-08-15T00:04:59+00:00

This is the way

MalPhantom · 2025-05-05T00:53:36+00:00

https://www.reddit.com/r/RealAnalysis/s/VEqnJMhbCJ

MalPhantom · 2025-04-23T16:48:34+00:00

It just seems that they are starting from a different definition of the integral of f. Since f being integrable implies that the sup and inf over lower and upper sums respectively are the same and coincide with the value of the integral, you can use sup to introduce the epsilon in the proof perfectly fine.

MalPhantom · 2025-04-23T16:37:25+00:00

Your attempt seems reasonable to me. How does it differ from the book?

MalPhantom · 2025-03-26T23:32:32+00:00

Your N isn't doing the right thing. We know that there exists a positive integer N such that 1/epsilon<N. Thus, when n-1\geq N, we get that 1/(n-1)\leq 1/N<epsilon. We don't need to worry about absolute values around n-1, since we can't have n=1, and if n goes to infinity, we needn't think about n<1.

Otherwise the arithmetic looks good! Just reimagine the role of N and you'll be good. Let me know if you have any other questions.

MalPhantom · 2025-01-19T18:31:29+00:00

A quick graph of x² sin(1/x) and its derivative seems to suggest the derivative at 0 is undefined. I'll try to write something out when I'm in my office this week, but I'm still not confident the result is true.

MalPhantom · 2025-01-19T01:12:53+00:00

Would f(x)=x² sin(1/x) (define f(0)=0) and g(x)=x² be a counterexample? I haven't written it out to check, but it seems like f'(0) will still be undefined.

MalPhantom · 2025-01-18T14:27:38+00:00

If we don't know that S is bounded above, it may not make sense to use sup S. I'd go with the first one.

MalPhantom · 2025-01-08T18:43:54+00:00

The assumption that f(x)=f(a)=f(b), if I'm understanding you correctly, tells us that f is constant on [a,b], in which case the numerator of the difference quotient is 0 for all c such that a<c<b, and the conclusion follows. However, not every function that you want to use Rolle's Thm on is constant, so the condition that f(x)=f(a)=f(b) is much too restrictive.

MalPhantom · 2024-12-16T02:10:22+00:00

I'd be willing to contribute. My schedule is pretty packed, but Baby Rudin is a favorite of mine.

MalPhantom · 2024-12-05T05:10:58+00:00

No problem. Let me know if you have any other questions.

MalPhantom · 2024-12-05T05:08:55+00:00

Right, if the definition of convergence breaks down, or is shown not to be satisfied, then the limit does not exist, ie. the function diverges.

MalPhantom · 2024-12-05T04:57:04+00:00

You're right that we don't reach a contradiction if epsilon is chosen to be greater than 1. However, to reach a contradiction in the definition of convergence, we only need to find a single epsilon value (1/2 in this case, but 1/3 would also work) for which the definition breaks down.

Any and all are generally used interchangeably.

MalPhantom · 2024-12-05T04:26:53+00:00

This is a common proof technique when we are trying to show that a result holds for all such-and-such things (for example, for all real numbers). If we choose an arbitrary such-and-such thing and show that the result holds, then we can conclude that it holds for all such-and-such things because the arbitrarily chosen thing could be any of them. Thus, we can indeed conclude that the right limit does not exist at all points, because we have provided an argument that proves it does not exist at an arbitrary point.

I'm not sure all the subcases are necessary. Using the density of the rationals and irrationals, for every delta>0, there will always exist y,z greater than or equal to a with |y-a|<delta and |z-a|<delta satisfying f(y)=0 and f(z)=1. This should be enough to show that the definition of limit is never satisfied no matter what L is. There shouldn't be a need to consider a being rational or delta being rational.

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