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Find the values of p for which the limit exists

  1. Oct 19, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the values of p for which the limit exists: lim (x-->0+) xp sin (1/x)

    2. Relevant equations

    None

    3. The attempt at a solution

    Alright so I've broken it down to 3 cases:

    p=0

    then we have lim (x-->0+) x0sin (1/x) = lim (x-->0+) sin (1/x)
    Let xn=1/nπ and yn=1/(2n + 1/2)π
    f(xn)= sin (1/xn) = sin (nπ) n=1,2,...,n
    lim (n-->∞) xn = 0
    lim f(xn) = 0

    I'm not too sure how to deduce that the limit doesn't exist here.

    p<0

    lim (x-->0+) xp sin (1/x)
    f(xn) = (1/nπ)p sin (nπ)
    f(yn) = (1/(2n + 1/2)π)p sin ((2n + 1/2)π)

    lim (n-->∞) xn=0
    lim (n-->∞) f(xn)=0

    We have lim (n-->∞) xn=lim (n-->∞) f(xn)=0 so the lim (x-->0+) xp sin (1/x)=0

    Not too sure about this conclusion either...

    p>0

    Using the squeezing theorem:

    -1<sin(1/x)<1
    -xp<xpsin(1/x)<xp

    So if we have:
    lim (x-->0+) xp=0
    lim (x-->0+) -xp=0

    Then, lim (x-->0+) xp=lim (x-->0+) -xp=lim (x-->0+) -xpsin(1/x)=0

    Thanks in advance!
     
  2. jcsd
  3. Oct 19, 2011 #2

    SammyS

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    Instead of the sequence used here, pick one that has odd number multiples of 1/(π/2), i.e., 1/(π/2), 3/(π/2), 5/(π/2), ...

     
  4. Oct 19, 2011 #3
    OK then I get lim (n-->inf) x_n=+inf and
    lim (n-->inf) f(x_n) =0. Is this enough to show that lim (x-->0+) sin (1/x) does not exist?
     
  5. Oct 19, 2011 #4

    SammyS

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    It's the sequence f(xn) that's of interest here.

    So, if [itex]\displaystyle x_n=\frac{1}{(2n-1)\frac{\pi}{2}}\,,[/itex] what is f(xn) ?
     
    Last edited: Oct 19, 2011
  6. Oct 19, 2011 #5
    f(x_n)=sin (pi n I pi/2), then I get lim (n-->inf) f(x_n)=+inf and lim (n-->inf) x_n =0
     
  7. Oct 19, 2011 #6

    SammyS

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    First look at 1/xn = (2n - 1)(π/2) = π/2, 3π/2, 5π/2, ... , for n = 1, 2, 3, ...

    Of course (xn)0 = 1 .

    So f(xx) = sin(π/2), sin(3π/2), sin(5π/2), ... = ? ? ?
     
  8. Oct 19, 2011 #7
    so f(xn) is not defined because it oscillates between 1 and -1?
     
  9. Oct 19, 2011 #8

    HallsofIvy

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    The sequence is certainly defined. It is the limit of the sequence that is not defined.
     
  10. Oct 19, 2011 #9
    What is unclear to me is how to use the sequence to prove that the limit doesn't exist.
     
  11. Oct 19, 2011 #10

    SammyS

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    Break the given sequence up into two sub-sequences; one with the odd terms, the other with the even terms.
     
  12. Oct 21, 2011 #11
    A brief visit at my professor's office clarified the problem. You pick two sequences xn and yn tending to 0. In the case p=0 case f(xn)=0 f(yn)=1 with xn=1/nπ and yn=1/(2n + 1/2)π.

    then lim (n→∞) xn = 0 ≠ 1 = lim (n→∞) yn.

    Therefore lim (x→0+) xp sin (1/x) does not exist for p=0.

    Thank you both for your help.
     
  13. Oct 21, 2011 #12

    HallsofIvy

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    Yes, that was exactly what SammyS said.
     
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