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Math Help - Prove that h is not continuous at 0

  1. #1
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    Prove that h is not continuous at 0

    Let L subset to R and define

    h(x) = {sin(1/x), x not = 0
    h(x) = { L, x = 0

    Prove that h is not continuous at 0. (Hint: Prove by contradiction,
    taking epsilon = 1/2 and use question 1)


    question 1:

    Suppose delta > 0.


    (a) Prove that there exists a positive integer n such that

    0 < 1/((4n+1)pi/2) < delta

    and sin ((4n+1)pi/2) = 1


    (b) Prove that there exists a positive integer m such that

    0 < 1/((4m+3)pi/2) < delta

    and sin ((4m+3)pi/2) = -1



    Can anyone help me with this, I appreaciate it
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  2. #2
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    Quote Originally Posted by 450081592 View Post
    Let L subset to R and define

    h(x) = {sin(1/x), x not = 0
    h(x) = { L, x = 0

    Prove that h is not continuous at 0. (Hint: Prove by contradiction,
    taking epsilon = 1/2 and use question 1)


    question 1:

    Suppose delta > 0.


    (a) Prove that there exists a positive integer n such that

    0 < 1/((4n+1)pi/2) < delta

    and sin ((4n+1)pi/2) = 1


    (b) Prove that there exists a positive integer m such that

    0 < 1/((4m+3)pi/2) < delta

    and sin ((4m+3)pi/2) = -1



    Can anyone help me with this, I appreaciate it
    Remember that to prove h is not continous at 0 we must show that there exist an \epsilon >0 such that for all \delta >0 there exists an x \in \mathbb{R} such that \vert x \vert < \delta and \vert L - \sin (1/x) \vert \geq \epsilon. As you're advised take \epsilon=1/2 and assume L \in [0, \infty) then by (b) of question 1 we have an x such that 0<x<\delta and \sin (1/x)=-1, now since L\geq 0 we have \vert L-\sin (1/x) \vert >1>1/2. If L \in (-\infty,0) use (a) and a similar reasoning
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