Epsilon -delta proof for sqrt function

Рет қаралды 10,540

Prime Newtons

5 ай бұрын

In this video, I showed how to prove the limit of a sqrt function using epsilon-delta

Пікірлер: 40

@anglaismoyen 5 ай бұрын

I think this is the fourth or fifth epsilon-delta video I've watched without fully understanding what it's all about. No problem, I'll keep trying until one day I get it.

@flamewings3224 5 ай бұрын

It’s about visualisation. Imagine the grahp of y = sqrt(x). And you wanna be sure that lim as x -> 4 you get y = 2. Cause sqrt(4) = 2, but… In the limit problem you HAVE TO be sure about values NEAR the point. So, there are delta, which is for near x values, and epsilon, which is near y values. And you wanna be sure that more delta gets smaller, than more your y values gets close to 4 with more epsilon gets smaller… Sorry for maybe not good explaining, but try to watch something about visual proof

@PrimeNewtons 5 ай бұрын

Great explanation 👌

@kingbeauregard 5 ай бұрын

Imagine a function, and a point on that function where you want to prove it's continuous. Now, can you imagine a rectangle centered at that point, of such dimensions that the function never hits the top or bottom edge? And can you shrink that rectangle down to nothing, such that the function never hits the top or bottom edge at any size? If you can do that, it means that the function truly is getting closer and closer to that point as you zoom in, all the way down to hitting that point. Soooo, epsilon-delta is about demonstrating, with mathematical rigor, that you can indeed construct a rectangle with the proper proportions to make this happen; if the rectangle exists, then the limit is proven. That's the basic concept, as it makes the most sense to me. The height of that rectangle is 2*epsilon, and the width of that rectangle is 2*delta; and there is a relationship between epsilon and delta such that, as epsilon gets smaller, so does delta.

@PrimeNewtons 5 ай бұрын

@@kingbeauregard How have you been? Welcome back!

@kingbeauregard 5 ай бұрын

@@PrimeNewtons Thanks! I've been busy with ... I don't even know what any more. It's like someone's been stealing hours out of my day, and I want them back. But it's good to see you again; you're looking well, and I approve.

@mariojr2585 3 ай бұрын

You are the best one. Watching you From Mozambique

@SiyaNyuli 2 ай бұрын

your hand writing is awesome

@mohammedaminelm7836 10 күн бұрын

I love your videos!

@DagmawiTesfaye-ev2tx 2 ай бұрын

Something close to genius You're one of the greats Has anyone ever told you that I mean you're great you really are We appreciate it. Where r u from

@PrimeNewtons 2 ай бұрын

Thank you! 🇳🇬

@kingbeauregard 5 ай бұрын

I like the cut of your jib sir; also your new hat. Also, I like the brutally methodical way you approach this: find some way to yank an |x - a| out, and then set the rest to the maximum value it can take in a strategically-chosen domain around x = a. Also, I like that you used the conjugate. I did not; I opted for a more general approach, which meant more work but got me to the same place. We know that t^n - b^n = (t-b)*(a bunch of terms), or equivalently, that (t-b) = (t^n - b^n)/(a bunch of terms). Now, suppose t = sqrt(x) and b = sqrt(a), and n = 2: then you can replace "sqrt(x) - sqrt(a)" with "(x - a) / (a bunch of terms)". At that point, I got to where you got to, but it took more work.

@LORDLDUQ 5 ай бұрын

Great video!!

@punditgi 5 ай бұрын

Brilliant! 🎉😊

@williampeters71 Ай бұрын

nice video to get he point over would state that if a < b the 1/a > 1/b

@danobro 5 ай бұрын

First comment that's not a bot, let's gooo!

@orey0721 3 ай бұрын

You're special one

@ryemiranda6800 5 ай бұрын

Do you have video suggestions or playlists where I can understand limits? Especially these epsilon delta proofs. I want to try to advance learn these lessons in calculus even Im in highschool learning algebra 2.

@zianiera 5 ай бұрын

Thats correct argumentation

@timothywesley1901 3 ай бұрын

What eraser does he use because his chalkboard always looks brand new

@frxysse 5 ай бұрын

Hi i'm just studying limit proofs at uni. It would be very helpful if you could do a epsilon-delta proof in 2 variables, thanks

@williampeters71 20 күн бұрын

listened again getting clearer we assume a delta less than 1 what if the limit of the function dne then this would be false

@NigusYilma-tj6dc Ай бұрын

Special one

@rivasu1030 5 ай бұрын

Hello, good explanation. But if you begin |x-4|

@shmuelzehavi4940 5 ай бұрын

Nice explanation, however I'm a little bit confused. Isn't it simpler to prove formally that: √x ⟶ 2 as x ⟶ 4 in the following way ? |√x - 2| = |(x - 4) / (√x + 2)| = |x - 4| / (√x + 2) ≤ |x - 4| / 2 Now, let ε > 0 . We define now: δ = 2ε . Therefore, for: 0 < |x - 4| < δ = 2ε we obtain: |√x - 2| ≤ |x - 4| / 2 < 2ε / 2 = ε Therefore: |√x - 2| < ε ∎

@DutchMathematician 5 ай бұрын

Your proof is almost 100% correct. The part that is missing is the fact that δ must be chosen in such a way that the restriction 0 < |x - 4| < δ ensures that the value of x belongs to the domain of √. Hence, δ can be at most 4. Prime Newtons chose 1 as an upper bound for δ but didn't mention the reason that this choice is valid (with respect to the domain of √).

@shmuelzehavi4940 5 ай бұрын

@@DutchMathematician You're right, and therefore we have to chose: δ = min {2ε , 4}.

@MichaelIfeco-tj1jc 21 күн бұрын

What if it's a cube root or even a fourth root

@JourneyThroughMath 5 ай бұрын

Maybe Im just not used to these problems or Im missing some tiny detail, but it was always problems like this that seemed very hand wavy

@kingbeauregard 5 ай бұрын

They're counterintuitive as heck, that's for sure. They come down to, if I can establish a relationship between epsilon and delta, then the function is continuous, and that feels like a non-sequitur. But we've established a particular relationship between epsilon and delta, that says something about, the closer you get to (a, L) vertically, the closer you get to (a, L) horizontally too.

@klementhajrullaj1222 4 ай бұрын

For me x=4 it's ok only for 1/(Vx+2)=1/4, because |x-4|

@glorrin 5 ай бұрын

I am unconvinced by the necessity of saying |1-4|=0 => 1/(sqrt(x)_2) |x-4|< 2 epsilon and choose 2 epsilon as our delta. Oh wait. Not all function are like sqrt(x) some do not have an obvious minimum/maximum this is a great too for any situation :)

@SimthandileMthe-ed5uq 2 ай бұрын

can you please explain why you say the square roots of(x)+2 is always positive?

@Totsy30 Ай бұрын

Probably because you cannot take the square root of a negative, and since there is no subtraction in that statement, it'll always be positive.

@hqs9585 4 ай бұрын

Why do you have to use value of 1, Just use the inequality |a+b|

@h1a8 5 ай бұрын

Here's a simple proof Let d=e>0 and 0 < |x-4| < d |x - 4| = | [sqrt(x)-2] * [sqrt(x)+2] | = | [sqrt(x)-2] | * | [sqrt(x)+2] | => | [sqrt(x)-2] | = |x-4| / | [sqrt(x)+ 2] | < |x-4| < d = e