Lecture 34: Limit Comparison Test for Improper Integrals of Type I and Type II.(see pinned comment)

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  • เผยแพร่เมื่อ 22 ธ.ค. 2024

ความคิดเห็น • 18

  • @saumitradeo5336
    @saumitradeo5336 3 ปีที่แล้ว +4

    Sir, In the first example, for DCT, if theta=5pi/6, then the inequality for the last two wouldn't hold.

    • @DrMathaholic
      @DrMathaholic  3 ปีที่แล้ว +3

      Yes, you are correct dude.. i was bit casual 😔 its only correct till 0 to π/2 and from π/2 to π inequality reverses...
      So the best way is substitute π-theta= x.

  • @nighatlone77
    @nighatlone77 3 ปีที่แล้ว +2

    The set [a, b] belogs to [c, d] is
    A) closed set
    B) open set.
    (C) nether closed nor an open set (D) neither an interval nor an open set

    • @DrMathaholic
      @DrMathaholic  3 ปีที่แล้ว +2

      Closed set

    • @nighatlone77
      @nighatlone77 3 ปีที่แล้ว +1

      @@DrMathaholic thank u sir....

  • @SumitGaming-
    @SumitGaming- 2 ปีที่แล้ว +1

    Sir in 3rd part 9:29 ,it converges

    • @DrMathaholic
      @DrMathaholic  2 ปีที่แล้ว +1

      How? 🤔🤔🤔
      1/x^3 is diverging right?

    • @SumitGaming-
      @SumitGaming- 2 ปีที่แล้ว +1

      @@DrMathaholic sorry .you are right

    • @DrMathaholic
      @DrMathaholic  2 ปีที่แล้ว +1

      @@SumitGaming- 🙌🙌🙂

  • @legendgaming8020
    @legendgaming8020 2 ปีที่แล้ว +1

    Sir how we solve this sum>>>>. 1/(1+x)x^1/2 converge or devearge i

    • @legendgaming8020
      @legendgaming8020 2 ปีที่แล้ว +1

      Limit x =0 to infinity

    • @DrMathaholic
      @DrMathaholic  2 ปีที่แล้ว +1

      rewriting as: root(x)/(x(x+1)) = root(x)(1/x - 1/(x+1)) = root(x)/x - root(x)/(x+1) .
      for first term, split the integration 0 to 1 and 1 to infinity. for 1 to infinity 1/root(x) diverges. so i think answer is divergent.

  • @helloworld-hv9oy
    @helloworld-hv9oy 3 ปีที่แล้ว +1

    Sir for the last questions is that test required ,beacuse we can directly evaluate that integral ?
    One more doubt ,it seems that for the limit test the convergence and divergence of f(x) depends upon the function g(x) we choose ...so can it be possible that there are two such functions and different answers come ?

    • @DrMathaholic
      @DrMathaholic  3 ปีที่แล้ว +1

      If you can evaluate directly then no problem at all.
      One can choose different different g(x) no problem in that, but the good thing is nature of f(x), whether it convergent or divergent, does not changes.

    • @helloworld-hv9oy
      @helloworld-hv9oy 3 ปีที่แล้ว +1

      @@DrMathaholic okay sir thank you !

  • @krishnadehrawala442
    @krishnadehrawala442 2 ปีที่แล้ว +1

    Sir we can apply any test from limit and direct ?

    • @DrMathaholic
      @DrMathaholic  2 ปีที่แล้ว +3

      Yes , just check the conditions..if the functions satisfies the conditions in that theorem then you can apply the respective test.
      If one test fails then try for second one..

    • @krishnadehrawala442
      @krishnadehrawala442 2 ปีที่แล้ว +1

      @@DrMathaholic ok