We cover several limit laws and prove them using epsilon delta proofs. email: mathjazz9@gmail.com website: almostcool.weebly.com facebook: / almostcoolnetwork
best explanation of limits properties proof on the internet i letterally searched everywhere
@kamalame6824
2 жыл бұрын
Literally I cant find words to thank you. I searched a lot in the internet to find the correct proof but this is the only source from where I got that. Thank you!
@ChristAliveForevermore
2 жыл бұрын
I think proofs by the limit definition are hard for people to grasp because of how non-intuitive it is *at face value.* It demands that we always choose some epsilon to be greater than 0 just to give the limit definition any logical sense at all (i.e., epsilon exists in the set of all Natural Numbers *at least* ). Then it demands that we *choose* some delta so that we can relate these two non-existent numbers by our actually-existing function. It's essentially a proof by induction where the assumed basis is epsilon, the assumed "kth value" is your chosen delta, and the "k+1 value" you want to show always exists is the limit itself (it is essentially a higher-order method of induction, which is *already* difficult for most people coming out of pre-Calculus to grasp immediately). The only difference is that in induction the function's kth value is *assumed to exist* because k is just a placeholder constant. Delta is *chosen* such that *your choice* determines if what we are trying to prove even exists. If you choose "wrong" (which is a foreign idea to pre-calculus) then it will *appear* as if the limit doesn't exist, but if you choose "correctly" (i.e., the value closest to x which corresponds to the interval of epsilon about f(x)), then almost by magic the limit exists *by definition.* Essentially *how we define a limit* is what determines if the limit exists, but, and this is the most important part, *because the limit was defined in a manner similar to how we intuitively understand the Principle of Induction (which is fundamentally a property of number that 'pops' out of the Natural Numbers), we feel we intuitively can grasp what a limit "is".* The kicker of all of this is that the limit only exists as a definition that we play with. Essentially there is no "infinity" or "continuity" in the Calculus, only limits and how define them. If I'm not mistaken, C.F. Gauss came to the same conclusion during his studies of the Calculus.
@file4318
Жыл бұрын
This comment has probably been the most useful comment I have ever read. I am currently self studying maths, and was really having trouble with this epsilon delta stuff. Since I hadn't studied proof by induction yet, your comment prompted me to quickly do so. Once I understood that, your comment clearly explained the links between a classic proof by induction and the epsilon delta proofs and then linked it back to a much more fundamental understanding of calculus. Thank you so much.
@user-ng6wl6dk1p
4 жыл бұрын
Thank you and your wonderful explanations!
@alial-musawi9898
6 жыл бұрын
I am so happy to come across this video. All the other explanations did not make sense. You explain things so well. Thank you for posting this video.
@mathjazz6930
6 жыл бұрын
Thanks!
@saipansmall
3 жыл бұрын
At 15:21, how do you get that |L*f(x)|=L^2/2 ?
@zhizhongpu8937
2 жыл бұрын
Clear explanation - thank you
@d.o.p2575
3 жыл бұрын
I have a bit of trouble understanding the multiplication proof. For |f(x)-l|
@gulagvo
2 жыл бұрын
there's no way i'm finally understanding it. ty so much, homie!!
@benclingenpeel8883
6 жыл бұрын
For the third proof (limx-->a cf(x) =cL) shouldn't it be |f(x) - L| < epsilon/(|c| + 1)? Seems to me that if you have just |c| in the denominator you would run the risk of dividing by zero.
@mathjazz6930
6 жыл бұрын
That is a good point. I should have stated that I am assuming that c is not zero, and I can get away with that because if c is zero then cf(x) is a constant, namely zero, and the limit of a constant is itself.
@ganakajayasinghe5375
4 жыл бұрын
It has to be |g(x)-M|
@elenaorins5284
4 жыл бұрын
So I understand that we have to manipulate for epsilon-delta, but for some reason, I do not understand if we are manipulating delta or epsilon. I just do not see it. In the video, you choosing some delta, but then you're making |f(x)-L|< e/some factor. I do not see how we are manipulating delta then
@samueltorres8928
4 жыл бұрын
As far as I understood, these existence proofs consist on finding that "delta" which allows us to imply the conclusion of the definition, and this process usually starts off by manipulating the right hand side of the inequality of the definition until we come across to a connecting expression with the hypothesis.
@amberkhan5959
5 жыл бұрын
Thank you
@mathjazz6930
5 жыл бұрын
Thank you!
@ravjesus4848
7 жыл бұрын
i don't understand why (|L-f(x)|)/(|Lf(x)|) < 2|f(x)-L|/(L^2) at 15:43
@mathjazz6930
7 жыл бұрын
First of all, thanks for making me look at this slide. The line that says "let delta be the larger of delta_1 and delta_2" should read "let delta be the smaller of delta_1 and delta_2." I would not have found that error had you not commented. Also, I should have split this into the case where L is positive and the case where L is negative, because there are a few subtle things that need to be addressed if L is negative that I did not handle fully in the video. I'll get around to fixing this eventually, but here is the answer to your question. Suppose L is positive. If |f(x)-L|
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