The limit of sin(1/x) as x approaches 0 does not exist (Proof) [ILIEKMATHPHYSICS]

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  • Опубликовано: 25 дек 2024

Комментарии • 7

  • @ansumanc
    @ansumanc 4 месяца назад +1

    We can also create two sequences (a_n and b_n), which tends to 0, but f(a_n) and f(b_n) not going to the same limit, thus proving the limit doesn't exist. One way of taking such sequences can just be to take the the numbers where f(x) is =1, and the other case where its -1, which we can solve for.

  • @saaah707
    @saaah707 5 месяцев назад +3

    Mind blown
    I need to get this book asap

    • @ansumanc
      @ansumanc 4 месяца назад

      I have it, its a very good book, along with Principles of Mathematical Analysis by Rudin

  • @draaagooonegggggg
    @draaagooonegggggg 5 месяцев назад +2

    beautiful proof!

  • @davidlawler706
    @davidlawler706 5 месяцев назад +5

    Can you do this proof using the negation of the definition of a limit?

  • @edcoad4930
    @edcoad4930 13 дней назад

    These proofs always do my head in. I was thinking what is it in this proof that means you can't apply the same logic to sin(x) limit doesn't exist by saying epsilon = 2 and setting x = k.pi or even setting epsilon = 1? By eliminating the "one over" everywhere gets the same....or is it the archimedian step that is actually key. Sometimes these videos need (possibly) stupid examples to show the value of these proofs.

  • @atticushatfield9948
    @atticushatfield9948 Месяц назад

    I love you