Prove the limit does not exist. Real Analysis Example (2.1)

[@praneelbhatnagar2333]

hello, where we get epsilon = 1/4 from?

[@garytiner508]

Basically, because 1/4 is the one-half the distance between the two possible function values (i.e., half of the distance between 1/2 and 1). We could have chosen any number for epsilon as long as 0 < epsilon < 1/4.

[@praneelbhatnagar2333]

@@garytiner508 hey thank you for your reply. I don't understand why we pick an epsilon between 0 and 1/4 whats the rule for picking and trick

[@anuj8855]

Thanks a lot sir you clear my very deep doubt 🙏🙏🙏🙏🙏🙏

[@sadececansu9]

Thanks a lot for this clear explanation😇😇😁

[@ashenafidejene9512]

great work , it is simple and clear

[@garytiner508]

Thank you for your nice comment!

[@360mathematics6]

Very nice explanation

[@slandrix5240]

One question, how do you know what epsilon to pick? Is it just for convenience?

[@jacobguerreso675]

Same question. Everything else makes perfect sense. Trying understand why epsilon equals greater than |f(x1)-L| wouldn’t work

[@jacobbarats1493]

Pretend that you didn't know it was 1/2 at the beginning. We just need to make the formula work for some epsilon, so at the end when we see that |...| + |...| > 1 when know that one of those terms is greater than 1/2 which we then let be the epsilon

[@jesusdanielmiguelcova2360]

Same question everything else is perfect

[@gustavstreicher4867]

The size of the jump discontinuity is 1/2. The halfway point between those two is your best average guess of what the limit could have been if it existed. So, the distance between that best guess and either of the actual values on the disconnected curves in the function is 1/4. Like others stated here, you could leave your choice of epsilon to the end. You just need to find a single choice of epsilon for which the rest is true to disprove the limit. So, choose whichever makes that the case for what you've worked out despite it.

[@soheil9255]

That was great . Thanks a lot .

[@yiokreverso]

If I have a more complex function, that F(x1) and F(x2)aren't numbers, like "lim x - > 2 of g(x) = { x^(2) if 0

[@garytiner508]

Sorry for incredibly slow reply. So the limit doesn't exist at x=2. Use epsilon =1/2 (or anything between 0 and 1/2). This is because 1/2 is one-half of the distance between lim x->2- f(x) = 4, and lim x->2+ f(x)=5. Then basically just follow my example.

[@TaharProd]

what a goat

[@TaharProd]

this is awesome

[@TaharProd]

thanks a lot

[@adesina1]

Thanks sir

[@namupalaaunenyanyukwenindeuten]

❤🔥🔥