In this video, we construct a function on domain (0,1) that has a limit at each point in (0,1) except at 1/2. We use the delta-epsilon definition of the limit to prove that the limit does not exist at 1/2.
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@anuj88553 жыл бұрын
Thanks a lot sir you clear my very deep doubt 🙏🙏🙏🙏🙏🙏
@ashenafidejene9512 Жыл бұрын
great work , it is simple and clear
@garytiner508 Жыл бұрын
Thank you for your nice comment!
@sadececns00 Жыл бұрын
Thanks a lot for this clear explanation😇😇😁
@360mathematics63 жыл бұрын
Very nice explanation
@metalliron88214 ай бұрын
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
@slandrix52403 жыл бұрын
One question, how do you know what epsilon to pick? Is it just for convenience?
@jacobguerreso6752 жыл бұрын
Same question. Everything else makes perfect sense. Trying understand why epsilon equals greater than |f(x1)-L| wouldn’t work
@jacobbarats14932 жыл бұрын
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