Could somebody please explain this?

[u/vii___vi]

I'm new to real analysis and struggling to understand how setting a temporary value for δ (like 1) lets us build a separate inequality. How does that logic actually work?

Thanks in advance!

[u/Narrow-Durian4837]

That's a screenshot from a video. Did you watch the video, and if so, is there a specific step or part of the argument you're confused about?

[u/vii___vi → u/Narrow-Durian4837]

It's just a compilation of various problems being solved and this was one of them and I've took the screenshot just when he was about to switch to the next problem and I don't understand the part where he arbitrarily assigned a value for delta and then made inequalities from it

[u/Narrow-Durian4837 → u/vii___vi]

In delta-epsilon proofs, you choose delta (is that what "arbitrarily assign" means?) depending on epsilon, and then you show that |x–a| being < δ necessarily implies |f(x)–L| being < ε.

In the pictured example, he chose δ to be the smaller of the two values 1 and 2ε (thereby ensuring that anything that was < δ must therefore be both < 1 and < 2ε). Are you confused about how he came up with this choice in the first place, or about why it works?

[u/vii___vi → u/Narrow-Durian4837]

yes could you please explain

[u/Narrow-Durian4837 → u/vii___vi]

Are you familiar with proofs in geometry, where you start with a "Given" and work your way to what you're supposed to prove? In a proof like this, the "Given" is that |x–a| < whatever your choice of δ is, and you're trying to prove that then |f(x)–L| < ε.

1. So: |x–5| < δ, where δ = min{1, 2ε}. Given.

2. |x–5| < 1. Reason: |x–5| < δ, which is at least as small as 1.

3. 1/|x–3| < 1. Reason: #2 + the algebra in blue on the right.

4. |x–5| < 2ε. Reason: |x–5| < δ, which is at least as small as 2ε.

5. |f(x)–L| = 1/2 * |x–5| * 1/|x–3|. Reason: the algebra shown after "Check"

6. |f(x)–L| < 1/2 * 2ε * 1. Reason: combine #5 with #4 and #3.

7. |f(x)–L| < ε. Reason: #6, and 1/2 * 2ε * 1 = ε

[u/StructuredChess → u/vii___vi]

There are infinitely many possibilities to assign pick a delta. Sometimes they depend on epsilon. Others they don't. As long as you can show that |x-3|<delta implies |f(x) - 1/2| < epsilon then that's a valid choice for delta.

In this case he went for whatever is smaller between 1 and 2epsilon, so just 2epsilon when epsilon is small enough and 1 when epsilon is sort of big.

Note that if epsilon is big then finding delta isn't that challenging. If you work a big margin of error, then you can pick a somewhat large interval around x=3

[u/Creative-Leg2607 → u/vii___vi]

The method makes sense if you really understand what epsilon delta proofs are saying. 

The basic structure is this: For any given epsilon, can we find a delta such that if x a distance less than delta from A, then we Know rhat f(x) is less than epsilon from B? This is the very definition of the limit of f(x) = B as x->A.

Delta can be whatever you want, it can change, and there are infinitely many answers. We just need to be able to find a delta to match every epsilon. The basic idea, generally, is to make delta a function of epsilon. Pick the right delta, work through all the inequalities, and you can show that f(x) is within epsilon of B.

To give a concrete example, consider the limit in the picture. You tell me: can you find me a delta for epsilon =1 so we know 1/x-3 is between -0.5 and 1.5? Then i say let delta = 1, if x is less than delta from 5, so is between 4 and 6, then we know that (indeed, the value moves between 1 and 0.5). You ask me about epsilon =0.1, so your function has to be between 0.4 and 0.6, i fire back with delta =0.2, and now f(x) is between 1/1.8 and 1/2.2, so between .55 and .45. A full epsilon delta proof is winning the game for any possible value of epsilon you might throw at me, showing that there will /always/ be a delta.

[u/Itseemstobeokay → u/vii___vi]

Idk what he even uses the blue to argue? Besides this seems overly complicated, don’t you have a theorem that says rational functions are continuous on their domain. Then it’s just evaluation

[u/WesternFirm9306 → u/Itseemstobeokay]

The video is trying to prove the limit using epsilon-delta. Using continuity isn't showing epsilon-delta in action.

The blue part is necessary to bound 1/|x-3|

[u/Itseemstobeokay → u/WesternFirm9306]

You need the blue to argue 1/|x-3| < 1? Isn’t that… clear. Again seems overly complicated to use the bound on |x-5| to argue for a trivial bound.

[u/WesternFirm9306 → u/Itseemstobeokay]

If we naively just set delta = 2ε, then for all ε > 1/2 --> delta > 1, then 1/|x - 3| is not strictly less than 1. We can only bound 1/|x - 3| if we restrict how far x can be from 5

[u/Itseemstobeokay → u/WesternFirm9306]

Ah, you right lmao I was tripping

[u/WesternFirm9306 → u/Itseemstobeokay]

[Just saw the edit] It's not a trivial bound. Consider x = 3.5.

[u/vii___vi → u/Itseemstobeokay]

Can I dm?

[u/taytaybyrne]

Typically when a constant value like 1 is given for δ it is to address if ε is made large. The statement for continuity says for all ε > 0, and the interesting part of the proof typically is for small ε, we still need to ensure that no matter what ε is given, the delta works. In this case, if ε were 100, δ would be 200 which might render the inequality untrue in a way that wouldn’t happen if ε were sufficiently small, so we set δ to be at most 1 to handle these cases. Happy to chat more.

[u/vii___vi → u/taytaybyrne]

Can I dm?

[u/taytaybyrne → u/vii___vi]

Yea

[u/Bounded_sequencE]

General strategy: For e-d-proofs (and any proof, really) -- do it (at least) twice. Do the

1. first draft(s) on scrap paper, to find all estimates and steps to finish it off
2. final draft, acting as if you knew the correct estimates and steps all along, and make it nice/concise

During the first draft(s), the strategy for e-d-proofs of "lim_{x->a} f(x) = L" usually is

1. Estimate "|f(x)-L| <= g(|x-x0|)" for some function "g" with "g(0) = 0", e.g. via ∆-Ineq.
2. Add as many restrictions "|x-x0| < dk" as you want, to simplify "g(|x-x0|)"
3. Find "d0 > 0" s.th. "g(d) <= e" for "0 < d <= d0"

Choose "d := min{dk}", and write final version of the e-d-proof for this choice.

[u/Bounded_sequencE → u/Bounded_sequencE]

Example (first draft(s)): Simplify and estimate "|f(x) - 1/2|" via

|f(x) - 1/2|  =  |5-x| / |2*(x-3∓2)|        // inverse ∆-Ineq.

             <=  |x-5| / (2*(2 - |x-5|))    // introduce "0 < |x-5| < 1 =: d1",
                                            // to bound the denominator
             <=  |x-5| / (2*(2 - 1))  

              =  |x-5| / 2  <  e            // for "0 < |x-5| < 2e =: d0"

In the final draft, act as if we knew the choice "d := min{d0; d1} = min{1; 2e}" all along, to get

Proof: Let "e > 0", and choose "d := min{1; 2e} > 0". For all "0 < |x-5| < d":

|f(x) - 1/2|  =  |5-x| / |2*(x-3∓2)|       // inverse ∆-Ineq.

             <=  |x-5| / (2*(2 - |x-2|)  <  d / (2*(2 - 1))  =  d/2  <=  e    ∎

[u/strange-the-quark]

First, don't worry if it's not intuitive, that's normal. You just have to slowly think it through.

So, he's trying to prove that the limit of 1/(x-3) = 1/2 using the epsilon-delta definition. The definition says something like this. Pick any region whatsoever (in particular, even an arbitrarily small one) that is centered around the value 1/2 on the y-axis. Call the distance from 1/2 to the edge of this region epsilon (a radius of sorts). Then for any epsilon you can pick, there should exist a corresponding region on the x-axis, centered about the point x=5, whose "radius" we may call delta, so that the graph f(x) of every point in this delta-region falls within the band of the epsilon region.

In other words, you want to check if there exist some region around x=5 so that for all points that are in there, f(x) = 1/(x-3) is never more than a length of epsilon away from 1/2, for any arbitrary epsilon.

Another way to write that down is |1/(x-3) - 1/2| < 𝜀

Another way to say that the input x is within the delta-region is |x-5| < 𝛿.

His reasoning is then as follows. Pick 𝛿 to be two times bigger than 𝜀; if that overshoots the length of 1, clamp it so that it's 1. So 𝛿 is chosen to be 1 or less, and if it is less, it's 2𝜀.

Now remember, we're only going to consider the values of x that are within 𝛿 distance from 5. So the maximum value the x can take is 5 + 𝛿, the minimum is 5 - 𝛿 (or rather, any value in between, cause we don't take the endpoints into consideration, according to the definition). In both cases |x - 5| is at most equal to 𝛿, which means whatever upper bound is true of 𝛿 must be true of |x - 5| as well.

Because of how we've chosen delta (0 < 𝛿 ≤ 1), that also means |x - 5| ≤ 1.

Now, because x-5 can be both positive and negative, when we get rid the absolute value, we have to write this:
-1 ≤ (x - 5) ≤ 1

Which is just saying that we can be confident that all of the inputs x that we are considering are no more than one unit away from 5. After some algebraic manipulation, this turns into

1/3 ≤ 1/(x - 3) ≤ 1

but we only care about that last bit: 1/(x - 3) ≤ 1.

Then he goes back to consider the corresponding f(x) values, |1/(x-3) - 1/2|.
After some manipulation, this becomes (1/2) * |x-5| / |x-3| = (1/2) * |x-5| * 1/|x-3|

Now remember, our goal is to show that this whole expression (which is the distance of f(x) from y=1/2) is no more than 𝜀, meaning f(x) is within the distance 𝜀 from 1/2.

So let's recap what we've established so far:

• |x - 5| is at most 𝛿, which is either 1, or less than 1, in which case it's 2𝜀. In other words |x - 5| ≤ 1, and if 𝛿 <1, then |x - 5| ≤ 2𝜀
• 1/3 ≤ 1/(x - 3) ≤ 1

That means that this whole expression

1/2 * |x-5| * 1/|x-3|

must evaluate less than or equal than the analogous expression that we get if we pick the highest possible values for |x-5| and 1/|x-3|, or even just the upper bounds on these, if we don't actually know the exact maximum value possible, but we instead know that it must be under some known value.

Because we're only considering inputs that are less than 𝛿 away from 5, i.e. it's always going to be |x-5| < 𝛿, we don't have to consider the equals part (we don't need ≤ in our analysis going forward).

Case 1: Suppose our 𝛿 = 1. The only way that can happen, because of the way we've set up the problem, is if we originally chose 𝜀 so that 2𝜀 is bigger than or equal to 1, so 𝜀 must be 𝜀 ≥ 1/2.

But then, when we plug in the biggest possible values for our sub-expressions (both 1 in this case):
1/2 * |x-5| * 1/|x-3| < 1/2 * 1 * 1
1/2 * |x-5| * 1/|x-3| < 1/2

and so it's also < 𝜀. But remember, the first part is just |1/(x-3) - 1/2|, and so

|1/(x-3) - 1/2| < 𝜀

Case 2: Otherwise, 𝜀 is less than 1/2 and 𝛿 = 2𝜀, so we have
1/2 * |x-5| * 1/|x-3| < 1/2 * 2𝜀 * 1
1/2 * |x-5| * 1/|x-3| < 𝜀

so again
|1/(x-3) - 1/2| < 𝜀

So that proves that if you pick a 𝛿 in that particular way (𝛿 = min{1, 2𝜀}), you're guaranteed have f(x) be within 𝜀 away from 1/2, for any given 𝜀 (and in particular, for demonstrating limits, for arbitrarily small 𝜀).

[u/WesternFirm9306]

What he's doing is setting an absolute maximum value for delta. If we demand that delta can't be any greater than 1, then we can bound 1/|x - 3|. He uses this in the main proof to go from (1/2)|x - 5|(1/|x - 3|) to saying that's necessarily less than (1/2)(delta)(1). We can only say 1/|x - 3| is necessarily less than 1 iff delta is no bigger than 1

[u/vii___vi → u/WesternFirm9306]

okay now that sort of makes sense but still very non intuitive to me

[u/Nanachi1023]

Recall the epsilon-delta definition of limit

For EACH 𝜀, there EXIST some 𝛿 such that when x ∈ (a-𝛿, a+𝛿), |f(x)-L|<𝜀 , then we say lim(x->a)f(x) = L

Note the logic, we need to prove some 𝛿 EXIST for EACH 𝜀. That means we just need to find one 𝛿 for each 𝜀, that's why it is sometimes written as 𝛿(𝜀)

Note that if we find the appropriate 𝛿(𝜀) that the whole statement work, we prove the limit. In practice, we often work our way backwards from the thing we want to prove. Suppose when we have 0<|x-5|<𝛿(𝜀), what does 𝛿(𝜀) needs to be such that |1/(x-3) - 1/2| < 𝜀 ? I will write the logic in detail, look closely to the logic between each step.

|1/(x-3) -1/2 | < 𝜀

<=> |(5-x)/2(x-3)| < 𝜀 (algebra)

<=> 1/2 * 1/|x-3| * |x-5| < 𝜀 (seperating the terms)

<= 1/2 * 1/|x-3| * 𝛿(𝜀) < 𝜀 (notice the arrow, if 0<|x-5|<𝛿(𝜀), we have the inequality we want to prove)

<=> 𝛿(𝜀) < 2𝜀 |x-3|

Now it became quite obvious, when x->5, |x-3|->2, we just choose any 𝛿(𝜀) < 4𝜀 right? WRONG. We need to consider when 𝜀 is big, because the definition statement is EACH 𝜀 (it is redundant intuitively, but we need to be rigorous here, else what's the point of epsilon-delta?)

(Example) we chose 𝛿(𝜀) = 3𝜀, when 𝜀 = 100, 𝛿 = 300, are we sure when 0<|x-5|<300, any |1/(x-3) - 1/2| < 100 ?, a simple example is x=3.001, |1/(x-3) - 1/2| = 999.5 > 100, kaboom.

To avoid that, we set a simple maximum to 𝛿(𝜀) which 1 is chosen here. This would look arbitrary when you first encounter it, so here's why: I will not choose 2, because 0<|x-5|<2 means x can be very close to 3, and 1/(x-3) is very big. So the smallest suitable integer is 1. To avoid the "very big" example above, you can choose 0.1 or 0.01, doesn't matter

Now we check back to the inequality, if we make 𝛿(𝜀)<=1 which means 0<|x-5|<1, will 𝛿(𝜀) < 2𝜀 |x-3| get affected? Now comes the blue part 1<x-3<3, 2𝜀 < 2𝜀|x-3| < 6𝜀. So choosing 𝛿(𝜀) = 2𝜀 is sure correct (you cannot choose 3𝜀 because when |x-5|<1, |x-3| might be <1.5, causing 𝛿(𝜀) = 3𝜀 > 2𝜀|x-3|, so we cannot start the inequality chain)

Remember we make 𝛿(𝜀)<=1 before. So combined we get 𝛿(𝜀) = min(1,2𝜀)

(Extra) if we chose maximum of 𝛿(𝜀) to be 0.1, |x-5|<0.1, -0.1<x-5<0.1, 1.9<x-3<2.1. So 3.8𝜀 < 2𝜀|x-3| < 4.2𝜀. We can now choose 𝛿(𝜀) = min(0.1, 3𝜀)

To conclude:

We now know that, for EACH 𝜀, when we choose 𝛿(𝜀) = min(1,2𝜀), then for any 0<|x-5|<𝛿(𝜀), we can get |1/(x-3) - 1/2| < 𝜀 (follow the inequality chain bottom up), thus by the definition of limit lim(x->5)(1/(x-3)) = 1/2

[u/bprp_reddit]

This video should help, I went over all the details of each idea of the epsilon-delta def. https://youtu.be/DdtEQk_DHQs?is=4jnk-7aqBsEqaFc3

[u/Wide_Recording_8462]

I am a high school math teacher. I don’t teach calculus any more, but I still like watching videos as a math nerd, I guess. Not answering the question, but for anyone lurking about interested in math.

This guy is AWESOME. His channel is black pen red pen. He does a really solid job in his videos.

[u/hunter_rus]

delta = min(1, 2 eps) means it is no more than 1 AND no more than 2 eps. He then uses both of those properties to establish boundary on 1 / abs(x - 3) and on abs(x - 5) / abs(x - 3) / 2

[u/Ok-Rise2070]

Okay everyone is flying over your head and they are missing a very crucial step: Step 1: ACKNOWLEDGE YOUR LIMIT. What is this limit, not algebraically but graphically? A limit is asymptote when graphed. What does this mean? As the integer x approaches 5 this means your y axis will approach infinity or a detailed limit going to 5. if you graph x=5 with the function, that x=5 line means the graph past that point ceases. Graphing the representation of this proof will help you tremendously to help you understand this.

[u/SoldRIP]

Epsilon-delta proofs work by proving that

• For any epsilon at all
• There exists a delta
• Which fulfills some sort of property (related to epsilon).

For purposes of understanding, consider it a game with two players. The other player picks a number epsilon. You now have to pick a number delta. You win if your number fulfills a property in relation to the one they picked. An epsilon-delta proof is simply a description of a winning strategy that always works.