Epsilon-Delta proofs: Can't we make the limit equal anything?

Epsilon-Delta definition explanation: • Epsilon Delta Limits i...
Proofs that use the delta-epsilon definition of the limit can be confusing because it seems like we can prove that the limit is anything we want if we pick the right value of delta. In this video we prove that limits are unique and go over some examples of disproving limits!
Calculus Problems playlist: • Calculus Problems
0:00 Background
3:46 Example 1
9:46 Example 2
15:45 Backward proof?
17:14 Proof of uniqueness
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Music: C418 - Pr Department

[@lucasfrykman5889]

I've always thought about this ever since my first analysis course. It frustrated me that I just had to accept it.

[@hybmnzz2658]

I really enjoyed the proof of limits being unique. I usually proceed with contradiction assuming L =/= M and show f(x) can't be bounded in two places at once. This is more elegant.

[@kingbeauregard]

Excellent video! It's good to see straightforward cases where epsilon-delta fails to cough up a successful relationship between epsilon and delta. This illustrates the point of the ordinary epsilon-delta process: you want to be able to say that, as epsilon goes to zero, there is always a corresponding delta, and the proof of it is that you were able to show math connecting epsilon to delta. I think of it like this. You have your function f(x), and you want to test that the limit of f(a) is L. So imagine a rectangle centered at (a, L); you want the rectangle to be of proportions such that the function never touches the top or bottom edges of the rectangle. Now, can you scale your rectangle all the way down to nothing, such that the function never touches the top or bottom of the rectangle at any size? If you can do that, then the limit at (a, L) exists. And if you think about all those rectangles and the function never touching the top or bottom, it's like you've drawn a big "X" centered at (a, L), and the function goes through the regions on the left and right of the "X", but never the regions on the top and bottom. What are the dimensions of that "X"? Well, the height and width correspond to epsilon and delta respectively.

[@MuPrimeMath]

I would be careful with the "big X" part, since that sounds a lot like Lipschitz continuity. Does that handle the case of the cube root of x being continuous at zero?

[@kingbeauregard]

@@MuPrimeMath Hmm, I see your point: the closer you get to x=0, the more the function resembles a vertical line. Dang if I can figure out how to epsilon-delta it though.

[@kingbeauregard]

... I'm ending up with delta = epsilon^3, but does that work? It can't be that simple, can it? You're right that my "X" model doesn't work, though. It's still about rectangles (I think), but those rectangles don't have to scale, just as long as both dimensions go to zero.

[@MuPrimeMath]

I think that delta=epsilon^3 works for the x=0 case! The thing to keep in mind is that the rectangles can change shape as they get smaller. For the cube root of x, the rectangles get more and more skinny as epsilon gets smaller.

[@kingbeauregard]

@@MuPrimeMath Thanks! We never covered epsilon-delta in my college years (somehow), and I've been struggling to make sense of it out of a sense of challenge. Your videos help a lot! Beard's still looking solid BTW. Keep it!

[@stickmanbattle997]

I’m not a math student but I want to learn this kind of stuff before going to college

[@tomkerruish2982]

Wouldn't it be simpler to just point out that R¹ is Hausdorff and therefore limits, when they exist, must be unique?😁

[@MuPrimeMath]

That's essentially what I did in the last section, using the fact that metric spaces are Hausdorff.

[@tomkerruish2982]

@@MuPrimeMath Well, that's what I get for making a smart-ass comment when I'm not even halfway through a video. Masterfully done!

[@AnakinSkywalker-zq6lm]

Is the μ a function or a constant?

[@ralfbodemann1542]

It's constantly functioning extremely well!

[@AnakinSkywalker-zq6lm]

@@ralfbodemann1542 Well that explains it!

[@devrajyaguru2271]

Is there any condition such as " epsilon is always greater than or equal to delta" for any value of delta?.....because the epsilon should converge more faster than delta, right?

[@MuPrimeMath]

There isn't necessarily any such inequality relation between epsilon and delta. For example, take the limit as x→0 of 2x, and then the limit as x→0 of x/2. In the first case we must have delta ≤ epsilon/2 (meaning epsilon CANNOT converge faster than delta), whereas in the second case we can take delta = 2*epsilon. As a side note, for any value of delta that works for a given epsilon, any positive number smaller than that delta also works!

[@pledab]

This thumbnail is so troll

[@alexx8126]

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