The math sorcerer: *understands the whole limit definition within epsilon and delta* Also the math sorcerer: * hmmm what is 9+6+4*
@shutupimlearning2 жыл бұрын
how is it possible for you to make this proof make sense???? Truly a Math Sorcerer
@maxpercer71194 жыл бұрын
I like this approach and it generalizes well, even with rational functions. The math sorcerer is actually using the triangle inequality, if we want to achieve full rigor. I will illustrate with a linear then a quadratic example. Show lim 2x + 3 = 5 as x→ 1. Given ε > 0 , there exists δ(e) such that for all x 0 < | x - 1 | < δ ⇒ | (2x + 3) - 5 | < ε But it is true that | (2x + 3) - 5 | = | 2x - 2 | = | 2 ( x - 1) | = 2 | x - 1 | < 2 δ. if we choose δ ≤ ε/2 the result follows . Quadratic example: Show lim x^2 -1 = 8 as x→ -3. Given ε > 0 , there exists δ(e) such that for all x 0 < | x + 3 | < δ ⇒ | (2x + 3) - 5 | < ε But | (x^2 -1 ) - 8 | = | x^2 - 9 | = | (x+3)(x-3) | = | x + 3 | * | x - 3 | < δ | x - 3 | = δ | x + (-3) |
@usctrojanfreak4 жыл бұрын
So you're saying I have to show work?
@ashveet420 Жыл бұрын
the intro though 🤣.Great video btw.
@niccoarcadia41793 жыл бұрын
Had to watch this twice but got it. Thanks!
@TheMathSorcerer3 жыл бұрын
Awesome !!
@user-qj3rv2mo1b10 ай бұрын
I loved how u forced down understanding at the lowest level into head so well. I enjoyed it and enjoy watching delta epsilon proofs. I request u to do many more delta epsilon videos on YT. Thank you
@gabrielcolombo68252 жыл бұрын
Really thanks man, your explanation finally helped me understand that proof
@shakirulislam24774 жыл бұрын
Freakin hell what a cool dude
@riyodjenero Жыл бұрын
Thank bro your help is very appreciated
@sarelaiber27755 жыл бұрын
Hi, great video! Much appreciated. Would love to see as many more advanced calc videos as you're inspired to add... 2 questions (contingent on my proper understanding - please correct me if I'm wrong): 1. I'ts okay to plug x=3 since for the domain 1
@GhostyOcean4 жыл бұрын
1. To make things easier to type, let g(x)=x²+2x+4. You want to find a bound for |g(x)|, and you know that 1
@GhostyOcean4 жыл бұрын
2. You don't need to reiterate that x≠2 because you already said that by saying 0
@raylittlerock39404 жыл бұрын
In the proof proper, you substitute definite values for x. But isn't the proof supposed to hold " for all x" ?
@gemacabero64823 жыл бұрын
I don't understand why delta can be epsilon over 19, because if it is then 19 times delta will not be smaller than epsilon, but instead it will be equal to.
@RikiFaridoke11 ай бұрын
I was amazing on your tricky step by step when you doing it right and correct, but you must realize that cubic function can be divide to quadratic equation
@cleopascletus27076 ай бұрын
Thanks the sorcerer
@AlexanderZim3 жыл бұрын
This is exactly what I needed. Thank you! PS You've got a great hat ;)
@martijn1303704 жыл бұрын
very instructive, thanks!
@TheMathSorcerer4 жыл бұрын
You're welcome!
@devnirwal44244 жыл бұрын
awesome
@TheMathSorcerer4 жыл бұрын
Thank you
@klementhajrullaj12226 ай бұрын
Can you prove it, that the limit when x goes to 2 of 2^x=4 with the language of epsilon and delta?
@p.c27504 жыл бұрын
@4:11 - but it is less than delta not equal to. So why replace it with < delta? I did not get that part, please break it down if you could
@joaohax523 жыл бұрын
Try with a numerical example: 3 < 5 3.4 < 5.4, it's clear
@p.c27503 жыл бұрын
@@joaohax52 thanks!
@mohfa1806 Жыл бұрын
But why we did not assume that delta =2 or 3.5 or.....etc??... Thx
@SomeOne-wb6np Жыл бұрын
.
@thenewdimension98323 жыл бұрын
Great ❤️❤️❤️❤️❤️
@kageyamatobio-h4q2 жыл бұрын
please can you make a video on how to proof the limit as x approaches 2 of the the function root of x