epsilon-delta proof ultimate study guide

Рет қаралды 97,481

Күн бұрын

We will prove 10 limits by using the epsilon-delta definition. Check out this related video: • college calculus class...
See the full 100 limits: • 100 calculus limits (f...
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0:00 81, epsilon-delta linear case
7:12 82, epsilon-delta linear case
11:09 83, epsilon-delta linear case
13:58 84, epsilon-delta sqrt(x) case
18:17 85, epsilon-delta sqaure root case
21:53 86, epsilon-delta quadratic case
28:22 87, epsilon-delta quadratic case
32:14 88, epsilon-delta x^3 case
38:27 89, epsilon-delta 1/x case
43:22 90, epsilon-delta 1/(2x+1) case
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Пікірлер: 118

@bprpcalculusbasics 9 ай бұрын

New video: Your first and second day in college calculus: kzfaq.info/get/bejne/fbCkgK6a2qfFdWQ.htmlsi=GTBLG_F3moYnhvJq

@batteryjuicy4231 8 ай бұрын

this man is an absolute legend. When he learned this stuff, there was no black pen red pen and when he learned it so well to create this channel to teach us bozzos

@IncendiaHL 8 ай бұрын

My mathematical analysis exam coming up, I realized how rusty and inexperienced I am with these. There are plenty of videos explaining the concepts, but that's not the hard part. The hard part is actually getting experience in using them, and this video!!! Is!! Exactly!!! What I wanted!! An incredible teacher just trucking through a boatload of Proofs. Thank you so much!

@annaholm9704 Жыл бұрын

Oh my God😃 This guy is a godsend. I'm so glad he made this video! I'm only in calc 1, and because of him, I already know the lectures before they are taught.

@kingbeauregard Жыл бұрын

Can I say a conceptual thing? Imagine you've got your function f(x), and there is a point (a, L) on the function and you want to prove that it's continuous. So, imagine a rectangle centered on (a, L), and it is tall enough that the function never touches the top or bottom edges of the rectangle. Now, can you resize the rectangle down to nothing, without the function touching the top or bottom edge at any resizing? That's what epsilon-delta is trying to do, abstractly rather than visually: the rectangle would be 2*delta wide and 2*epsilon high. So you're just working out the proportions of the rectangles; if you can derive proportions for the rectangles, the function is continuous.

@iteoluwaoladejo4240 Жыл бұрын

@@kingbeauregard W concept. I never thought of it that way. Thanks a lot for sharing.

@josefmakili4062 Жыл бұрын

To dopa nena

@leseagleexe6947 9 ай бұрын

Youre an absolute gem sir

@bprpcalculusbasics 9 ай бұрын

Thanks!!

@luciekiss1618 Жыл бұрын

Thank you for your passion ! This video is so helpful. Keep doing what youre doing.

@jacobc6452 Жыл бұрын

This is very helpful! I am in intro to proofs this semester but also in an analysis class that is graduate level so it has been challenging so far. This really makes things much more understandable - thank you!

@omerdogancanbilgili7401 Жыл бұрын

bro you literally save my academic career thank you

@eddiephysics 10 ай бұрын

Condensed and extremely easy to follow along!

@del3t3d Жыл бұрын

Wonderful, exam tomorrow and you simplify things so that they can be easily well understood :)

@Akane_Kana 9 ай бұрын

This video explained everything. Love that!

@Antonwas12 5 ай бұрын

Thank you so much, at first I didn't understand anything but because of your explanations I was able to solve the last 3 by myself

@antipsikiyatriKizi 7 ай бұрын

Best video about this topic on KZfaq.

@emmecuh Жыл бұрын

Great video as always, would love to see some one sided and infinite limit proofs someday.

@alibinnaseer Ай бұрын

thank you soo much!!! I watched sal's video, then yours and then again sal's video and then went to ChatGPT and then came back to this, rewatched it and now i finally understood it !!! May Allah bless you !! 💖💖

@user-oh4tf8lj9y Жыл бұрын

you are amaizng toutr i hope you are my proffessor instead of my profferssor THANK YOU A LOT

@user-vr2ny2ff9u Жыл бұрын

You are outstanding, thanks you, i understood this !! In my institution no one explained this...

@SuperYoonHo Жыл бұрын

Thank you so much!

@magnetic_monopole Жыл бұрын

This helps me a lot!

@kiri0725 4 ай бұрын

Thank you very much !!!

@macdaveshub Жыл бұрын

I love your teaching sir 🥰🥰 This is a topic I have been finding soo difficult to understand. But you have elaborated it completely. And I have learnt it. Keep up the good work ✌️🥰

@mecidgaliba2854 Жыл бұрын

Thanks a lot for this comprehensive video. I was just wondering if you could make a video about how to write down this proof in exams. I mean not by using arrows or red markers. Thanks in advance!

@user-qy9py4lv9e 9 ай бұрын

Loved itt!

@Sdw_Bld 6 ай бұрын

How is this guy such a good teacher?!

@jeloabmelak9783 Жыл бұрын

Really helped me save the semister🎉❤

@stewartgarikayi4560 Жыл бұрын

Thank you Professor

@nvapisces7011 Жыл бұрын

It would be great if you could make a video on a guide to prove one sided limits as well as infinite limits and a mix of both. Then make separate videos proving the different limit laws (sum, product, quotient, power, root law)

@rain-uw9ju Жыл бұрын

isn't epsilon delta the method used to prove all of the different limit laws? I thought it was the basis for limits in general.

@TazzD83 Жыл бұрын

lucky that I found you!! 😁Thanks a bunch!

@user-qj3rv2mo1b 6 ай бұрын

At the end of each proof u prove it by deducing δ to get ε. It definitely makes sense, And grateful to you for wonderful explanation and so many interesting examples. Now I feel comfortable with proving limits using your GCSC process. However, I have also seen few other proofs, they determine δ like u do and then back track through the key controllable factors to the start, to show |F(x)-L|

@ishanschneider9289 13 күн бұрын

absolute legend

@waldomonroy452 Жыл бұрын

Right on time

@bhavishaysharma8648 Жыл бұрын

Okay. This was crazy!!! If you read this, thanks a lot bud :)

@bprpcalculusbasics Жыл бұрын

Thank you!

@NWS390 10 ай бұрын

It's grate sir❤

@danni4451 11 ай бұрын

God, I learned a lot from this video. Thank you so much! You teached so well. I couldn't even understand how this actually works, and after watching your video it makes me 恍然大悟.😁😁😁😁😁😁

@bprpcalculusbasics 11 ай бұрын

Glad to help!! 😃

@Music-zx4po 10 ай бұрын

Thank you for your work! I could follow your first example, however why exactly may one introduce the epsilon - what exactly permits this step? This is probably like proofing the premises.

@tarundeeparora865 Жыл бұрын

love you

@evanssmith1327 Жыл бұрын

Man please do you have moment of forces channel??? You are just good

@kingbeauregard Жыл бұрын

I notice a pretty common practical methodology on each of these, almost a road map for what to do with the math: 1) You start with "f(x) - f(a)"; find some way to mingle those two terms, such that an "x-a" emerges in there somewhere. 2) Separate the "x-a" from the rest of the expression, so you have something like "(x-a)*(other stuff)". That "x-a" will become our delta. 3) Get rid of any instances of "x" that appear in "other stuff".

@mikesmithyes967 Жыл бұрын

Do Measure Theory practice problems.

@yakobbeyene8764 11 ай бұрын

Thank you for your help! Would you proof this question by using epsilon and delta definition. Limit x approaches 1 ( 5x^5+4x^3+x^2+x+1=12

@ngozikanwachukwu6623 Жыл бұрын

honestly, I really don't like the epsilon delta proof at all. But thank you for making it easier and tolerable for me❣

@velmagiah4310 Жыл бұрын

Amazing

@thomasjefferson6225 Жыл бұрын

Helped me in analysis a lot.

@creepsow1236 Жыл бұрын

I love the shirt

@monsieurlambertz7706 Жыл бұрын

I'm sort of confused. When he says a/b0, it's all fine. But I've been thinking what if a=b? If a=b then a/b = 1 < a, so in this case a must be greater than 1. In other words, can I just assume a/b < a in all cases? Even when |x-a| = sqrt(x) + sqrt(a)?

@Drum11craze 11 ай бұрын

Can you do a video showing a proof of a limit as x goes to infinity? Can't find anything on this subject.

@alpasamudra3479 Ай бұрын

sir please do calculus 3 multvariable case

@user-bd8gv6po5w 10 ай бұрын

Appreciate that a lot sir

@Paparazzi1e 10 ай бұрын

This is cool 😎

@dearlymats428 8 ай бұрын

The most humorous lecturer in my life 😅😅😅😅😅😅wow I enjoyed the lecture , best lecture in my life ,you make it easier to understand with jokes # if you want to be adventurous put something like pi or root of 17 got me offf😅😅😅😅😅etc

@ronycb7168 26 күн бұрын

21:00 we could just get rid of only the square root like you had shown in another video, I just want to see if this is also correct

@cunningham.s_law Жыл бұрын

thanks can you do one where you prove inf and sup of a bunch of sets

@energy-tunes Жыл бұрын

by the supremum axiom all finite subsets of real numbers have a supremum. then you can easily prove that they also have infimum given this axiom

@sihlec1047 9 ай бұрын

24:19 I'am dead... use pi... use 1/e

@MicaelSLopes Жыл бұрын

Nice algebra class

@heroasik5423 8 ай бұрын

in limsqrt(x),x----->4 delta=4epsilon

@KIYAFey Жыл бұрын

what if as x goes to infinitive instead of constant number

@Professor-Q Жыл бұрын

I love you

@ryanfox6260 9 ай бұрын

I've been exploring a different technique perhaps you can comment on. I would set up the equation 4+e = 6(x+d)+2 and solve for d = e/6. I visualize it as setting one of the bounds on the y-axis equal to the outer bound on the x-axis. In general it's L + e = f(x+d), where you know values of L and x. thoughts?

@stevenstone7101 7 ай бұрын

Can you explain further?

@markobavdek9450 Жыл бұрын

Torture of mind with these letters. Anyway I'll watch this...

@Yt-ff6hn Жыл бұрын

Right said ...

@kingbeauregard Жыл бұрын

What if we were doing f(x) = e^x? Here's how I think it goes, but I'll show you where I run into trouble: |e^x - e^a| < epsilon e^a*|e^(x-a) - 1| < epsilon |e^(x-a) -1| < epsilon/e^a -epsilon/e^a < e^(x-a) - 1 < epsilon/e^a 1 - epsilon/e^a < e^(x-a) < 1 + epsilon/e^a Next step is to take the ln, which I can do because ln(x) is an increasing function just as e(x) is increasing. In other words, it's not going to flip the sign. BUT -- you can't take the ln of a negative, which means that epsilon must be less than e^a. Right? This is ultimately my question, but I'll continue the arithmetic. ln (1 - epsilon/e^a) < x - a < ln(1 + epsilon/e^a) Now we want to put this back in absolute value form: |x-a| < ln(1 + epsilon/e^a) ... and I took the "plus" term rather than the "minus" term because its magnitude is smaller. Finally, we have: delta = ln(1 + epsilon/e^a) In broad strokes, this holds up: since epsilon is always positive, the ln() is always positive; and when epsilon equals 0, delta also equals 0. But again, the epsilon-delta definition specifies "for ANY epsilon", and I'm only allowing 0 < epsilon < e^a. So, is this an invalid proof, because I'm specifying a maximum epsilon?

@dalek1099 Жыл бұрын

If mod(e^x-e^a)< epsilon then it is also less than epsilon' where epsilon'> epsilon so the maximum epsilon isn't a problem.

@tahastm2522 8 ай бұрын

hello thanks for your video can you help me to solve lim(x^1/2+1)x=1 =1?

@leonardobarrera2816 Жыл бұрын

Thanks If you don’t want to do the problem that I am solving for x^(x+1)=10 Don’t do it, I will se how I can solve it But thanks about all your videos Loves and prayers from Ecuador

@bprpcalculusbasics Жыл бұрын

I don't think that's doable algebraically. We can use Newton's method, or graphical method. Here's the answer on WFA www.wolframalpha.com/input?i=solve+x%5E%28x%2B1%29%3D10

@leonardobarrera2816 Жыл бұрын

@@bprpcalculusbasics Thanks a lot for respond my comment, I don't know, but I would like to go to the event that will be in North California, because I love maths, and also you will be there, and it will be great for all your subcribers that also will be in there, and I would like to subscribe for the event. Thanks a lot for the feedback Remember, you are genious, I would like to be like you, I guess, if the things go in the right way, I will be like you!!! Cheers!! Loves and players from Ecuador!!!!!!!

@SuperYoonHo Жыл бұрын

@@leonardobarrera2816 u soUnd like sir premath LOL

@leonardobarrera2816 Жыл бұрын

@@SuperYoonHo really???? I express like that because I didn’t now how to say like we say in Spanish!! Thanks, I would like to be in the tour mente of north California (Steve send in the channel of blackpenredpen)

@SuperYoonHo Жыл бұрын

@@leonardobarrera2816 Oooh that's cool! 👍👍

@niteshgiri8369 Жыл бұрын

lim x -> - 2 (x ^ 2 + 3x) = - 2 solve it by delta epsilon definition

@paulpitters2185 11 ай бұрын

You’re so smart

@retro_cody_ Жыл бұрын

This is really cool. So it’s basically mathematical manipulation 🤣. Finishing up calc1. I somewhat get the proofs but ofc need some practice and help :)

@heroasik5423 8 ай бұрын

graphically study the epsilon delta you will get simple epsilon/delta=f'(a)

@Firefly256 Жыл бұрын

What should I do if I can’t get the |x-a|? For example, [limit of 2x+1 as x->1] = 3 0 < |x - 1| < delta |2x - 1 - 3| = 2|x - 2| |x - 1| < 1 -1 < x - 1 < 1 -2 < x - 2 < 0 |x - 2| < 2 So I can only get to: 2|x - 2| < 4 I don’t have delta in the above, how should I approach this?

@fretzyjanebares2257 Жыл бұрын

Hello, I think you have incorrectly written I f(x) - L | in your proof. You wrote |2x - 1 - 3| instead of |2x + 1 - 3|. The 3rd line should be |2x + 1- 3| = | 2x -2 | = 2 | x-1 | < 2 * delta. Hence delta is epsilon/2. Hope this helps. 😀

@Firefly256 Жыл бұрын

@@fretzyjanebares2257 oh right 😅, so does this situation never appear at all for other functions?

@alexei2985 Жыл бұрын

At 45:50 you write down 1.7 as if it would technically work, then you suggest to use 1 instead because it is easier. I’m confused why 1.7 would technically work because when I use 1.7, I get a different answer at the end. Why would 1.7 work if it gives us a different answer?

@giana.2381 10 ай бұрын

The epsilon-delta definition of a limit only requires us to prove that a delta (such that the conditions are satisfied) exists. It doesn't require us to find all delta that satisfies the conditions. It also doesn't require us to prove that the delta is unique. So having a different delta is fine, as long as the conditions are satisfied. Sorry, this might not be useful to you anymore since I'm replying to you a couple of months late. 😅

@salvadorsabateromero9707 2 ай бұрын

21:40

@salvadorsabateromero9707 2 ай бұрын

13:48

@kaistrandskov Жыл бұрын

This is a little confusing for me. I prefer to do this: 1. We need to figure out how to choose a delta, when given an epsilon. So start with |F(x)-L| < epsilon, and plug in all the numbers for F(x) and L. This is what we know because we are given an epsilon. 2. Next, figure out how to choose a delta based on epsilon. Now, get this inequality into the form of |x-a| < G(epsilon), where G is some junk involving epsilon that we just solved for. Since our goal is to choose a delta so that |x-a| < delta, we can just choose a delta = G(epsilon). 3. Now go through the proof from the beginning, and choose a delta = the G(epsilon) we found earlier.

@salvadorsabateromero9707 Ай бұрын

32:43

@user-kh6cm1kd6y 8 ай бұрын

14:40

@energy-tunes Жыл бұрын

were you recording this for hours you sound like youre going to pass out in the entire second part

@jtlbb2 7 ай бұрын

Something doesn’t make sense. Around the 40 minute mark you had (x-2/2x) and you split it up into (x-2)(1/2x). Previously in similar problems you said to ignore the denominator because the numerator was greater than the whole fraction. Why is this different? After all (x-2/2x)< (x-2), so why is this different than problems you did earlier?

@bprpcalculusbasics 7 ай бұрын

Could u please give the exact time stamps?

@jtlbb2 7 ай бұрын

Of course. At 16:00 and then 39:09. I’m confused why the approaches are different.

@jtlbb2 7 ай бұрын

@@bprpcalculusbasics I think I might’ve figured out why. Is it because in the problem at 39:09, you can’t be sure the denominator is greater than 1? For example, the bottom could be 2x = 2(0.25) making it less than one meaning the fraction is larger than just the numerator. I think I just figured out the difference and didn’t think it through enough before asking.

@bprpcalculusbasics 7 ай бұрын

@@jtlbb2 Yes, exactly! If it was a 2+x on the bottom at 39:09, then we could also ignore the denominator. Also, if you would like to see other examples with x->inf or the limit being inf, this video is more complete: kzfaq.info/get/bejne/d8yiobyEmrmchnk.htmlsi=V8rMFSOiXrsenyz6

@jtlbb2 7 ай бұрын

@@bprpcalculusbasics thank you. I subscribed because you’re a very good teacher