limits with epsilon-delta definition! (linear, square root, and quadratic examples)

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@nyinakuemmanuel8164 3 жыл бұрын

I've been watching you videos since high school. I'm a Ghanaian offering BSc Statistics at University of Cape Coast. Am in my three-year. Thanks very much red pen black pen. This semester I'll do Advanced Calculus. So I'll watching every single of your videos on epsilon _delta proofs🙏🙏🙏👌💗

@zyriv2427 3 жыл бұрын

As a freshman in college, when I first started studying real analysis, I got confused by how to choose correctly ε in relation with δ. But this definitely helped me, thanks !

@vinicus508 10 ай бұрын

In fact its the other way arround. You choose δ in relation with ε.

@joefuentes2977 7 ай бұрын

In fact they are both related to each other... @@vinicus508

@dgilmore589 Жыл бұрын

This video was a lifesaver. Thanks.

@russellhan3534 Жыл бұрын

legend saving my grades...

@jujoropo 2 жыл бұрын

Wooow thank you very much! You help so much! Love your videos

@kingbeauregard 3 жыл бұрын

The part about arbitrarily choosing the delta has always been the most confusing part for me, because it seems like a nonsensical step. Here's my concept of what it's about. When you've got that expression on the left and epsilon on the right, what you're trying to do is come up with increasingly simpler expressions that are always bigger* than your original function; if you can finally get to a bigger* simpler expression that you know converges to your limit, then so must your original function. The problem is, it's pretty rare that you can come up with a simpler expression that is bigger* for all values of x. So that's when you cheat: you say "well I know that suchandsuch expression will be bigger* if I keep x within a certain narrow region". Like, think of y = e^x, and the limit at (0, 1). It would be nice to come up with some straight line that is always bigger* than e^x; then you could just say that, since the line converges to (0, 1), so must e^x. But there is no straight line that will do that for every value of x. So instead you cheat: you can say that, if we're limiting our x values to the region from -1 to 1, then for that region, a line with a slope of (e-1) that goes through (0, 1) will always be bigger* than the function. Once we make that tradeoff, we can proceed. *: When I say "bigger", I mean a function that is further away from the limit than our original function for every value of x in question. It's a squishy word but it makes instinctive sense I think.

@saurabhmaskara8666 3 жыл бұрын

Jesus christ bro, idk if u will ever see this but thank you so much for your explanation, my brain literally couldn't wrap itself around this step because it seems so arbitrary!

@echohasbinokiller4 3 жыл бұрын

A lot of times in mathematical proofs, some steps do seem to come out of nowhere. But a lot of that is due to the claim you're trying to prove itself. For example, if we claim that the limit of a function exists, then we have to prove the existence of that delta. So, by us choosing a delta so that we get the expression that we want, we manage to show by that line of reasoning that the claim is true and is in accordance with the definition of a limit. Pulling out that delta out of thin air is a useful way to do prove it because you already have the foreknowledge that the claim is true. Epsilon-delta proofs tend to be pretty difficult for many math majors first starting out with proofs. It's a leap in exercise of logic and I think much more should be done in the university curriculum to expose students to mathematical proof earlier and with more gentle approaches.

@kingbeauregard 3 жыл бұрын

@@saurabhmaskara8666 Thank you! I have spent the past month (possibly two) just trying to understand epsilon-delta, and once I got it (I think?), I figured out how I would explain it such that it made sense to a newcomer. I did a long-ish post about epsilon-delta over here, it might be worth looking at: kzbin.info/www/bejne/epXXdoShlKl7h9U I fully believe that epsilon-delta is confusing only because people aren't given an explanation of the process and the techniques before applying them. If you give people a "map" of what's going to happen with epsilon-delta, it's pretty comprehensible.

@robertveith6383 3 жыл бұрын

@@saurabhmaskara8666 Stop taking the Lord's name in vain and write in English words, please.

@AeonTrespassField 2 жыл бұрын

@@robertveith6383 Please stop trying to force your religious beliefs onto others, we don't consent.

@bautistaromaniuk4996 2 жыл бұрын

Thank you for the vids! They are super helpful. However, I still do not understand why we can simply ignore the √10-3x +4

@npdamh 2 жыл бұрын

because √10-3x +4 > 1, 3|x+2|/√10-3x +4 is less than 3|x+2|/1. When you compare 2 fractions with the same numerator, the one with the greater denominator is less than the other

@bisratmulugeta5718 Жыл бұрын

Thank you for the videos , on the previous video when you did root(2x+6)=4 as x approach 5 when solving you came to 2(x-5)/root(2x+6) +4 and you ignored the root and kept 4, making the delta 2/4. Why did you keep the 4 in the above question but not on this video?

@ozan_alkan Жыл бұрын

@@bisratmulugeta5718++

@Oliyadtsegaye Жыл бұрын

You are life saver 🙌 thank you mr

@dominicgrew4000 3 жыл бұрын

Thank you very much....great video as always

@Manase-q3k 4 ай бұрын

This was amazing ❤

@CHEESYhairyGASH Жыл бұрын

At 7:22, why do you ignore the integer in the denominator? In a previous video titled "how to easily write the epsilon-delta proofs for limits" you only ignored the square root part and kept the integer in the denominator (this happened at 6:33 in the other video). Why do you use different methods in these two very similar cases???

@yag1z560 3 ай бұрын

this is exactly why i am confused +++++++

@kimi9572 2 ай бұрын

Yeah i think he is wrong because the denominator can be any number besides 0. What you need to do is input -2 in the denominator and conclude that it cannot be bigger than 8. So, you put 8 in the denominator and the sign changes into "less than equal to". The delta value would be (8/3)epsilon.

@aliciapark523 2 жыл бұрын

damn. this is awesome!! THANK YOU!

@jerkersverkersson1401 Жыл бұрын

tack bror ❤

@marylamb6063 Жыл бұрын

One reason why you want to "tie" the value of delta with the value of epsilon is that you want to prove the limit of the function for ALL values of x. There will always be only one value of delta that will be true for only one value of epsilon.

@AlessioVragnaz98 3 жыл бұрын

Given Choose Suppose Check

@onenutwonder 3 жыл бұрын

I have no idea how or why the epsilon delta proofs work to show the limit is true but maybe that's why i find them to be so damn cool.

@kingbeauregard 3 жыл бұрын

You start with a big complicated function that you want to prove converges to (a, L). Now maybe you can't do much to prove a limit with the original function, but if you can come up with a simpler function that you know is always "bigger" than the original function and also converges to (a, L), then the original function must converge too. So the epsilon-delta process is about swapping out terms in the original function with terms such that you 1) make the entire expression "bigger" than the original function, and 2) also make it simpler. The best outcome is if you can find a straight line that is always "bigger" than the original function (within a specific region close to x=a anyway); that straight line will converge to (a, L), and therefore so will the original function.

@kushaldey3003 3 жыл бұрын

I need a little help in the last example, when we need to choose δ=min(1,ε/15). We could have said δ=min(1.5,__). Then, |x-2| 0.5

@MrDerpinati 3 жыл бұрын

yeah i think its just 1 is the most simple

@kingbeauregard 3 жыл бұрын

Yes. What is happening here, I think, is that we have realized that we can't solve the problem for every value of x, but we can solve it if we declare that we are considering only x-values within a given distance from x=a. The region near the limit is what interests us, so that is perfectly acceptable. Note that our choice of region will impact our final definition of epsilon.

@tzonic8655 3 жыл бұрын

@@kingbeauregard you opened my eyes

@yoyoezzijr 2 жыл бұрын

For the second one, wouldnt delta = epsilon also work? because the denominator √(10-3x) + 4 is always greater than 3, so the expression is always less than |x+2| which is less than delta which is equal to epsilon

@nikokoro5862 2 жыл бұрын

It does. In fact, you can pick any number on the interval (0,4] since "sqrt(10-3x) + 4" is always greater than or equal to 4. You're essentially just adjusting the region of your delta while remaining within the bounds of epsilon.

@jafecc 6 ай бұрын

What if we have two square roots adding together or subtracting one from another in the limits? Is it harder to prove using epsilon-delta definition? Please let us know.

@PodcastClips910 9 ай бұрын

thank you!!!

@danjalsoltani8654 2 жыл бұрын

i love ur videos

@judedavis92 3 жыл бұрын

For the delta inequalities, you don't actually need the 0 < part because we're dealing with absolutes?

@azarath3512 2 жыл бұрын

We need the 0 < inequality because the proof of the limit relies on the fact that you choose an x whose distance from the limit (in the x world) (denoted |x - a| in the definition of a limit) is not 0 nor greater than delta, which is the input distance that corresponds to the output distance that gets smaller and smaller for values of the input closer to the limit (in other words, the proof shows how you can have any x closer and closer to the value x approaches in the limit-but not equal to it-and still get a defined value of the output-or f(x)-that gets closer and closer to the value of the limit itself; the condition of "but not equal to it" is expressed in the "0