Delta Epsilon Proof with Cubic Function

[@djalalmaster1018]

The math sorcerer: *understands the whole limit definition within epsilon and delta*
Also the math sorcerer: * hmmm what is 9+6+4*

[@shutupimlearning]

how is it possible for you to make this proof make sense???? Truly a Math Sorcerer

[@maxpercer7119]

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) |

[@usctrojanfreak]

So you're saying I have to show work?

[@niccoarcadia4179]

Had to watch this twice but got it. Thanks!

[@TheMathSorcerer]

Awesome !!

[@WritersDigest-b8f]

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

[@gabrielcolombo6825]

Really thanks man, your explanation finally helped me understand that proof

[@ashveet420]

the intro though 🤣.Great video btw.

[@riyodjenero]

Thank bro your help is very appreciated

[@shakirulislam2477]

Freakin hell what a cool dude

[@RikiFaridoke]

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

[@sarelaiber2775]

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

[@GhostyOcean]

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

[@GhostyOcean]

2. You don't need to reiterate that x≠2 because you already said that by saying
0

[@raylittlerock3940]

In the proof proper, you substitute definite values for x. But isn't the proof supposed to hold " for all x" ?

[@gemacabero6482]

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.

[@martijn130370]

very instructive, thanks!

[@TheMathSorcerer]

You're welcome!

[@klementhajrullaj1222]

Can you prove it, that the limit when x goes to 2 of 2^x=4 with the language of epsilon and delta?

[@AlexanderZim]

This is exactly what I needed. Thank you! PS You've got a great hat ;)

[@kageyamatobio-h4q]

please can you make a video on how to proof the limit as x approaches 2 of the the function root of x

[@cleopascletus2707]

Thanks the sorcerer

[@mohfa1806]

But why we did not assume that delta =2 or 3.5 or.....etc??...
Thx

[@SomeOne-wb6np]

.

[@devnirwal4424]

awesome

[@TheMathSorcerer]

Thank you

[@p.c2750]

@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

[@joaohax52]

Try with a numerical example:
3 < 5
3.4 < 5.4, it's clear

[@p.c2750]

@@joaohax52 thanks!

[@thenewdimension9832]

Great ❤️❤️❤️❤️❤️