Epsilon-delta proof for rational functions

Рет қаралды 10,169

Prime Newtons

3 ай бұрын

In this video, I showed how to do an epsilon- delta proof for rational function

Пікірлер: 37

@spacetimemalleable7718 10 күн бұрын

An EXCELLENT explanation! One big positive is you went slowly and not rushed through it as most profs do. Thank you.

@baileydwyer453 3 ай бұрын

As someone who is starting to learn epsilon delta proofs at uni, this is gold.

@kingbeauregard 3 ай бұрын

For my money, Prime Newtons is the best epsilon-delta-er on all of KZfaq. Pay close attention to the brutal, almost terrifying, efficiency with which he turns the expression into |x - 3|*|(some function of x)|. Once you've gotten to that form, all you have to do is figure out the maximum value that "some function of x" can take. Most of the time, you have to say, "well, if we restrict ourselves to the domain from x=2 to x=4, the maximum value is ...". Or maybe it'll work better if you go from x=2.5 to x=3.5. It's up to you to pick a domain, just some narrow region around x=3 that makes calculations easy. (Make sure you stay the heck away from x=-1, because you smack into a singularity there, and that just wrecks everything.) The concept is, imagine you're drawing a rectangle centered at (3, -1) that is tall enough that the function never hits the top or bottom edge. Now, can you shrink that rectangle down to nothing -- as in, zero height and zero width -- such that the function still never hits the top or bottom edge? If you can do that, it means that, the closer you get to y = -1, the closer you also get to x = 3, so the function really does converge. In other words, if you can demonstrate mathematically that such shrinking rectangles exist - with a width of 2*delta and a height of 2*epsilon - then the function really is continuous at that point. Note that there's no one solution to these things: depending upon the exact math you perform, you might come up with "delta = 3*epsilon" while I come up with "delta = 4*epsilon". That's fine; ANY valid relationship between delta and epsilon will do.

@antonmilius5197 Ай бұрын

👋🏼

@omaryarali7805 2 ай бұрын

Lots of respect and greetings from Azerbaijan, sir.

@CANALIMG 3 ай бұрын

Thank you teacher! I was struggling so much with this topic at uni 😭

@JourneyThroughMath 3 ай бұрын

I love this channel and will forever be grateful for it. I will throw out one arguementitive point. He seems to believe that the finding of the delta constitutes the proof. That is not the case for all calc one teachers. (Name drop) Michael Penn teaches that from delta to epsilon is the proof and the finding of the delta is "scratch work". My point is, be warey of what your teacher expects. Prime Newton does always say that, it does not seem like he is saying that in this case.

@PrimeNewtons 3 ай бұрын

When I took Advanced Calculus, my professor made it clear that finding a delta was the hard part. Showing the delta works is just reverse engineering. So, anyone who can find a delta should be able to go from delta to epsilon. So I agree with Michael Penn. Delta to epsilon is the proof but not the hard part.

@SamsonMulugeta-w9t 14 күн бұрын

I hope u will also teach my son in the near future

@crunch_and_crunch Ай бұрын

nice work proff

@elifelif9395 Ай бұрын

You're amazing!

@algorithms_hub 20 күн бұрын

I like you man keep it up.

@Nkosinathi0025 3 ай бұрын

am in uni and this is so helpful plzz do more videos regarding maths because I find it hard to understand my lecture during class😭

@vincentmudimeli4430 2 ай бұрын

hi man iam still lost can you do more of this examples especially finding max and min for delta

@dannieee333 3 ай бұрын

thank you so much !!

@loganeliott6590 3 ай бұрын

You Didn't loose me 😊

@XxBiduxxX 3 ай бұрын

Never stop learning

@powercables1611 3 ай бұрын

I prefer to just fill in delta later, and just consider |f(x)-L| then try to bound it from above. For example here |f(x)-L|=2|x-3|/|x+1|, i can also say, |x-3| is something I can control, so all I need to worry about is the denominator, then can I bound it from below? And so on. We are doing the same thing here but I think conceptually this will be better in the long run, because for tougher examples, say proving lhopitals rule, the same idea builds intuition, but had we worked backward for our scratch it would have been less intuitive to work out the proof

@naturallyinterested7569 3 ай бұрын

Shouldn't you have picked 2/5 as the approx.? If we know A*B < e with A, B positive and we know 2/5 < A < 2/3, then wouldn't the only logically sound one be 2/5 * B < e, as 2/3 * B might overshoot? Imagine in the limiting case equality A * B = e and A = 2/4, then 2/4 * B = e < 2/3 * B. Or am I missing something?

@PrimeNewtons 3 ай бұрын

Think this way I claim 4x < y If it is true that 3x < y , can I say because of this, my original claim is true? Now consider changing 3 to 5. What do you think?

@naturallyinterested7569 3 ай бұрын

@@PrimeNewtonsOh, so I understand, thanks for clearing that up!

@Obliviousovertimer17 27 күн бұрын

@@naturallyinterested7569 it still made no sense if it is true that 2x < 5 then 3x might be bigger so i tend to agree with your first approach

@MichaelIfeco-tj1jc 13 күн бұрын

Please solve this "the limit as x tends to 1/3 3x-1/5x+1=0"

@catnip2906 Ай бұрын

Pay dirt and gold in understanding. The other level.

@teofeluskanime01 Ай бұрын

Am failing for how you got 3

@pianoplayer123able 22 күн бұрын

He did not add +3 but added +4 so he could get directly to x+1. So 2+1

@BartBuzz 3 ай бұрын

It seems that the proof would have been the same if you had picked "5" instead of "3" in the inequality 3

@PrimeNewtons 3 ай бұрын

Not the same. If 3 < x and 3 < 10, can you say x < 10 ?

@Armytechrex 3 ай бұрын

Ill never understand this definition, never made sense to me

@dennisrichards6596 23 күн бұрын

Does anyone see a parallel between the two goats and the the covering cherubs before God?

@griffinf8469 3 ай бұрын

I’ve never understood why we have to use the Epsilon-Delta proof to prove a limit. You prove the limit by doing basic mathematical calculations. Here’s my proof of the limit shown in the video: you have to show that the limit exists using 3^+ and 3^-. 3^+ simply means a number slightly bigger than 3, say 3.01. Plug 3.01 into the function and you’ll get approximately -0.995. Then, 3^- means a number slightly smaller than 3, say 2.99. Plug 2.99 into the function and you’ll get approximately -1.005. As you can see, both the left and right side of 3 converge or approach -1. Therefore, the limit exists and the answer is -1. This is all you need to prove if a limit exists or not. You don’t have to use the Epsilon-Delta proof.

@kingbeauregard 3 ай бұрын

The point of the epsilon-delta is, how do you know that there is no value at which the function does something weird? Like, how do you know that the function behaves as expected at 3.00000003928? You can try the value, sure, but then there are an infinite number of values to try, and that gets impractical. So epsilon delta operates by setting up a zone around your point where two things hold: 1) inside of that zone, you're less than a given vertical distance away from the point; 2) every smaller zone you set up, the same thing happens but with an even smaller vertical distance. If those two things hold, then the function has no choice but to converge as our intuition tells us it should.

@kingbeauregard 3 ай бұрын

@@davidgagen9856 YOU DON'T THINK it does anything weird. Now prove it.

@griffinf8469 3 ай бұрын

@@kingbeauregard Thank you so much! That makes a lot of sense!

@urgjendevetak3605 3 күн бұрын

my guy is still stuck in high school, either that or he's a physicist