Calculus 2.2 The Limit of a Function

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  • Опубликовано: 7 янв 2025

Комментарии • 17

  • @chrisencelan4021
    @chrisencelan4021 4 месяца назад +1

    Thank you so much Asher!

  • @vinmushwana9391
    @vinmushwana9391 3 года назад +4

    these videos are amazing. Keep up the great work

  • @xpertg91
    @xpertg91 Год назад +3

    We are not allowed a calculator so can you do one without a calculator

  • @tracya.bellian1159
    @tracya.bellian1159 2 года назад

    at 10:12 , i was wondering why do you say the limiting value of x/x =1? can you not use x=0.9, x=0.1, x=0.01 etc.? as long as the denominator is not zero the function still is defined right?
    thank u!

    • @asherroberts
      @asherroberts  2 года назад +1

      As long as the denominator is not 0, x/x evaluated at any value for x will be 1, since any nonzero real number divided by itself is 1.

  • @ltahoe9257
    @ltahoe9257 2 года назад

    When you graph the Heaviside function @ 18:25 you start at T is less that zero and plot it to the left of zero on the x axis, why do you plot T is greater than or equal to zero on the y axis?

    • @asherroberts
      @asherroberts  2 года назад +1

      In the picture I drew y is the value of the Heaviside function H. Since H(t)=0 for t

  • @BLAKMAN874
    @BLAKMAN874 2 года назад

    why do you express example #4 as sinx*pi and not sin(pi/x)? at 11:16

    • @asherroberts
      @asherroberts  2 года назад +1

      I express it as sin(pi/x). Each of the fraction inputs gets flipped when dividing, so the division turns into multiplication.

  • @vontellerpingham4676
    @vontellerpingham4676 3 месяца назад

    for example 5, why did you only investigate the values that approached 0 when x > 0 instead of doing x < 0 as well? I understand negative and positive does not matter for cosine functions, however there is an x^3 that changes depending on whether or not x is negative. I tested what the limits approached for x and the negative counterpart, when far from 0 the values are different, however when approaching close to 0 the values are the same whether or not x is negative. However, they are still not the same, is there a reason you did not explore both sides?

    • @vontellerpingham4676
      @vontellerpingham4676 3 месяца назад

      maybe it is because the value of x^3 when there is a really small number would just get exponentially smaller to the point it doesnt make a significant change to cos5x/10000?

    • @asherroberts
      @asherroberts  3 месяца назад

      @@vontellerpingham4676 You should really check values for x < 0 as well, but the book does not do so for this problem. Presumably, Stewart just wanted to stress the idea that sometimes tables can be misleading when trying to determine limits, and felt that this was sufficient to illustrate that.

  • @TurboRC101
    @TurboRC101 Год назад

    34:40 If you don't mind, I don't understand why you only graphed cosx . If tan is equal to sinx over cosx why choose the denominator and only graph cosx?

    • @asherroberts
      @asherroberts  Год назад +1

      Vertical asymptotes can occur when the denominator is 0, so we graph the denominator cosx to see where it is 0.

  • @sarahq8128
    @sarahq8128 3 года назад +1

    Thank you

  • @bokangfalatsi3736
    @bokangfalatsi3736 Год назад

    Thank you sir

  • @pawanyadav3399
    @pawanyadav3399 2 года назад

    Nuce