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Proving limit doesn't exist

  1. May 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Limit problem.png
    I have to show that the following function does not have a limit as (x,y) approaches (0,0)

    3. The attempt at a solution
    I tried taking different paths for example y=x or y=0 and switching to polar coordinates, but I don't get anywhere.
     
  2. jcsd
  3. May 28, 2015 #2

    pasmith

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    Consider approaching the origin along the line [itex]x = 0[/itex].
     
  4. May 28, 2015 #3
    Right so the first term cancels and I am left with: y sin ( 1/ y2 ) + sin ( 1 / y2 ) .
    So what do I conclude by taking the limit of this?
    y sin ( 1/ y2 ) vanishes since y goes to 0. I am a bit confused on what to do with sin ( 1/y2).
    Can I claim that limit doesn't exist because sin ( 1/ y2 ) alternates as y goes to zero and therefore doesn't have unique answer ?
     
  5. May 28, 2015 #4

    pasmith

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    How would you justify that rigorously?
     
  6. May 28, 2015 #5

    Zondrina

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    You can explain the dense oscillations near the origin by recalling ##\text{sin}(y) = 0## for ##y = n \pi, n \in \mathbb{Z}##.

    So ##\text{sin} \left( \frac{1}{y} \right) = 0## for ##y = \frac{1}{n \pi}, n \in \mathbb{Z}, n \neq 0##.

    Therefore we can say ##\text{sin} \left(\frac{1}{y^2} \right) = 0## for ##y = \frac{1}{\sqrt{n \pi}}, n \in \mathbb{Z}, n \neq 0##.

    There is a dense population of these points near zero; think about the interval ##(0, \frac{1}{\sqrt{\pi}}]##.

    As for showing it formally, you're going to have to squeeze some effort out, and you might need a sandwich to have enough energy.
     
  7. May 28, 2015 #6
    Would it be okay to take sequence an = √2/√nπ . Then this sequence obviously converges to zero as n goes to infinity. But f(an) alternates between 1,0,-1,0,...
     
    Last edited: May 28, 2015
  8. May 28, 2015 #7

    Zondrina

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    There is infinitely many points approaching zero:

    $$\frac{1}{\sqrt{\pi}}, \frac{1}{\sqrt{2 \pi}}, \frac{1}{\sqrt{3\pi}}, \frac{1}{\sqrt{4\pi}}, \frac{1}{\sqrt{5 \pi}}, \frac{1}{\sqrt{6 \pi}}, ...$$

    Group these into triplets:

    $$\left( \frac{1}{\sqrt{\pi}}, \frac{1}{\sqrt{2 \pi}}, \frac{1}{\sqrt{3\pi}} \right), \left( \frac{1}{\sqrt{4\pi}}, \frac{1}{\sqrt{5 \pi}}, \frac{1}{\sqrt{6 \pi}} \right), ...$$

    Notice for each triplet of points inside ##(0, \frac{1}{\sqrt{\pi}}]##, the function ##\text{sin} \left(\frac{1}{y^2} \right)## will oscillate from ##0## to ##-1##, then to ##0##, then to ##+1## and then back to ##0##.

    The triplets need not be ordered as I've shown. You can group different triplets and the function will oscillate in the same fashion, but potentially in a different order. For example, using:

    $$\left( \frac{1}{\sqrt{2 \pi}}, \frac{1}{\sqrt{3\pi}}, \frac{1}{\sqrt{4\pi}} \right)$$

    Will produce a different sequence for the oscillation (0,1,0,-1,0), but the behavior will be the same for each triplet. Namely, ##\left| \text{sin} \left(\frac{1}{y^2} \right) \right| \leq 1##.
     
  9. May 29, 2015 #8
    Simply speaking, we choose epsilon=0.9, we cannot get a corresponding delta to satisfy the epsilon-delta definition of limit.
     
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