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) |
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
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
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
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.
The math sorcerer: *understands the whole limit definition within epsilon and delta*
Also the math sorcerer: * hmmm what is 9+6+4*
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) |
So you're saying I have to show work?
how is it possible for you to make this proof make sense???? Truly a Math Sorcerer
Had to watch this twice but got it. Thanks!
Awesome !!
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
Really thanks man, your explanation finally helped me understand that proof
the intro though 🤣.Great video btw.
Thank bro your help is very appreciated
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
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
2. You don't need to reiterate that x≠2 because you already said that by saying
0
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
Freakin hell what a cool dude
This is exactly what I needed. Thank you! PS You've got a great hat ;)
Thanks the sorcerer
very instructive, thanks!
You're welcome!
In the proof proper, you substitute definite values for x. But isn't the proof supposed to hold " for all x" ?
Can you prove it, that the limit when x goes to 2 of 2^x=4 with the language of epsilon and delta?
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.
awesome
Thank you
please can you make a video on how to proof the limit as x approaches 2 of the the function root of x
But why we did not assume that delta =2 or 3.5 or.....etc??...
Thx
.
@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
Try with a numerical example:
3 < 5
3.4 < 5.4, it's clear
@@joaohax52 thanks!
Great ❤️❤️❤️❤️❤️