Epsilon Delta Continuity (Example 4): x^3

[@amandatran4207]

YOURE A GENIUS DUDE! YOU MADE A POINT I WAS STRUGGLING WITH CLICK FOR X^4. THANK YOUUU

[@drpeyam]

Yay glad I could help!!

[@frozenmoon998]

Watches a video and goes to the front page for other recommended videos: *Dr P uploaded 1 minute ago*

[@hemonben9004]

What a coincidence another continuous proof from de peyam as i enter youtube

[@s.kmithamo908]

PEYAM!!!!!!Some people whom should be considered as of great help towards attaining my degree

[@jamesbentonticer4706]

Another great video. Thanks for the daily content.

[@nebiyouyismaw8111]

Shouldn't you pick \delta < min{1, eps/c} instead of =?

[@nourdinespen8568]

Great explaining ❤

[@maqsudxorazm]

Thank you, Dr!

[@jidrit999]

why have you taken |x-x_0| < 1 first ? shouldnt epsilon be set first ? its like setting N for sequence and then deciding the band in which sequence of elements fall into

[@maahirbharadia7487]

Awesome vid really clear thanks ☺️

[@ShaolinMonkster]

very good , would like some conceptual/geometrical videos about the same example

[@arunsahoo3145]

sir can you make a video on hypergeometric series

[@josemanuelramirezgomez6206]

Thank you for this video, it help me a lot

[@zainabsidiq9403]

So what's the value of C when f(x)=x^3 is continuous at x=2

[@kurthayes2743]

plug in 2 for Xnot

[@Duedme]

6:29 Awsome one

[@tomascernansky3960]

thank you, thank you

[@purim_sakamoto]

ε-δごにゃごにゃなるので、こんな低いレベルまでフォローしてくれてありがてえ
1回生の初回の授業がきっとこんなんなんやろな～
レジ打ちより面白かっただろうなぁ

[@Moramany]

Yo isn’t C a Lipschitz constant in this case?

[@mcqueen424]

Correct me if I’m wrong but no because C depends on x_0. x_0 is fixed in our case but Lipschitz doesn’t deal with fixed points.

[@goofvos]

But what if C=0 ??

[@drpeyam]

Easier, |x|^3 < epsilon means |x| < cube root of epsilon, so let delta = cube root epsilon

[@goofvos]

@@drpeyam yeah but what if C=(|x²|+|x*x_0|+|x_0²|)=0 (bc x and x_0 can both be 0), then your delta is epsilon divided by zero, which is not possible.

[@drpeyam]

Still works, my constant has a 1+|x0| which is always nonzero

[@goofvos]

@@drpeyam ah ok thank you

[@bobjohnson1407]

good comment

[@talesamaral3744]

What if |x-x_0| >= 1?

[@floriankubiak7313]

Then it's greater than the delta we chose. Checking for continuity at a point distant from x_0 doesn't really make sense, hence you can assume an upper bound, but no arbitrary lower bound.

[@nitrojuliet]

love that