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Need help proving limit does not exist!

  1. Sep 20, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to prove a limit does not exist for f(x,y) as (x,y) -> (0,0). I already tried approaching from y=mx and I got the limit 0.


    2. Relevant equations
    f(x,y) = ((|x|^a)(|y|^b))/((|x|^c)+(|y|^d)) where a,b,c,d are positive numbers and a/c+b/d ≤
    1

    3. The attempt at a solution
    I got limit = 0 approached from y=mx.
    I can't use Squeez Theorem either. How should I approach this question? Just a hint would be good. Thanks!


    edit: Sorry, missed a + sign in the denominator!
     
    Last edited: Sep 20, 2012
  2. jcsd
  3. Sep 20, 2012 #2

    SammyS

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    Hello mathie_geek. Welcome to PF!

    Edited:
    [STRIKE]I assume your function really is:[/STRIKE]

    I see you corrected it to be:

    [itex]\displaystyle f(x,y)=\frac{|x|^a\,|y|^b}{|x|^c+|y|^d}\ .[/itex]
     
    Last edited: Sep 20, 2012
  4. Sep 20, 2012 #3
    Thanks!
    Trying to learn how to input functions :P
    I missed a + so denominator is |x|^c + |y|^d
     
  5. Sep 20, 2012 #4

    SammyS

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    Also, what is the restriction on a/c & b/d?

    Is it as you said, [itex]\displaystyle \frac{a}{c}+\frac{b}{d}<1\,,[/itex]

    [STRIKE]or is it[/STRIKE] [itex]\displaystyle \frac{a}{c}<1\ \text{ and }\ \frac{b}{d}<1\ ?[/itex]

    Added in Edit:

    Never mind, I see that it must be as originally stated.
     
    Last edited: Sep 20, 2012
  6. Sep 20, 2012 #5
    It's (a/c)+(b/d) <= 1
    can anyone help
     
  7. Sep 20, 2012 #6

    jbunniii

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    Hint: try setting x = t^p, where p is some power, and y = t^q, where q is some other power.
     
  8. Sep 20, 2012 #7
    I got lim y->0 ((|t|^ap)(|y|^b))/((|t|^pc)+|y|^d)) = 0
    and lim x->0 ((|x|^a)(|t|^bq)/((|x|^c)+(|t|^dq)) = 0

    Did I do something wrong?
     
  9. Sep 20, 2012 #8

    HallsofIvy

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    Are there conditions on a, b, c, and d? If a+b is not larger than c+d you will NOT get "0" as limit for y=mx.
     
  10. Sep 20, 2012 #9

    jbunniii

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    No, I meant that you should pick appropriate values for p and q (in terms of a, b, c, and d) which will give you a counterexample. Hint: try to choose values which will get rid of the exponents in the denominator.

    This will transform the problem into a limit involving a single variable, t. Then let that variable t approach zero. By doing this, x and y will approach (0,0) along a certain curve, instead of a straight line. If you do it right, you can get a nonzero limit.
     
  11. Sep 20, 2012 #10
    ok, I tried x = t^(d/c) and y = t^(c/d) as two different limits so I got:
    lim y->0 (|t|^ad/c)(|y|^b)/((|t|^d)+(|y|^d))
    lim x->0 ((|x|^a)(|t|^bc/d))/((|x|^c)+(|t|^c))
    those both =0 though.
    I'm sorry I'm not that good at this.
    Do you mean to put both of the substitutions into one limit of t->0?
    a/c <= 1-(b/d)
     
  12. Sep 20, 2012 #11

    jbunniii

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    No, try to choose powers which will make the exponent in both terms of the denominator equal to 1.

    Yes, make both substitutions at once, and then take the limit as t->0.
     
  13. Sep 20, 2012 #12

    SammyS

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    m_geek,

    Try letting (x, y) → (0, 0) along the path [itex]\displaystyle y=x^{c/d}\ .[/itex]

    That makes the denominator be fairly well behaved. I haven't checked it beyond that.
     
  14. Sep 20, 2012 #13
    OH, I let y= t^(1/d) and x=t^(1/c)
    so that lim t->0 (|t|^a/c+b/d-1)/2
    and since a/c+b/d-1 <= 0,
    then the limit would be 1/2 if a/c+b/d-1=0 or does not exist if a/c+b/d-1< 0.
    since the two limits do not agree, limit d.n.e.
    Is this correct? :D
     
  15. Sep 20, 2012 #14
    I tried that but I don't know why it didn't work out for me because I kept getting 0 as the limit..
     
  16. Sep 20, 2012 #15

    SammyS

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    Let's see your work for this, because it did work out for me & the limit is not zero.
     
  17. Sep 20, 2012 #16
    I let ##y = x^(c/d)##
    so that it simplifies to ##(|x|^ (a+(bc/d)))/2(|x|^c)##
    which as x->0 limit becomes 0/2 ? I'm not sure how to take into account the inequality of a,b,c,d though
     
  18. Sep 20, 2012 #17

    SammyS

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    I have fixed your Latex above, and repeat it below, so you can see the code.
    [itex](|x|^{a+(bc/d)}) /(2(|x|^c))[/itex]


    Now as a more readable fraction:

    [itex]\displaystyle \frac{|x|^{a+(bc/d)}}{2(|x|^c)}[/itex]

    Write that expression using |x| only once.

    Remember, [itex]\displaystyle \frac{x^R}{x^T}=x^{R-T}\ .[/itex]
     
  19. Sep 21, 2012 #18
    I got this after: [itex](|x|^{a+(bc/d)-c}) /{2}[/itex]
    expanding the exponents I have [itex]\displaystyle \frac{acd+bc^2-c^2d}{cd}[/itex]
    and given [itex]\displaystyle \frac{ad+bc-cd}{cd}<=0[/itex] which is 1/c of the exponent..
    So then |x| has an exponent negative so limit d.n.e right? cause limit of 1/|x| as x->0 d.n.e
     
  20. Sep 21, 2012 #19

    SammyS

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    Actually, that's [itex]\displaystyle \frac{ad+bc-cd}{d}\le 0[/itex]

    What if [itex]\displaystyle \frac{ad+bc-cd}{d}=0\ ?[/itex]


    Now, maybe you can adjust the path so that the exponent is always zero.
     
  21. Sep 21, 2012 #20
    Oh, I see. Since [itex]\displaystyle \frac{ad+bc-cd}{d}> \displaystyle \frac{ad+bc-cd}{cd}[/itex] then does that mean it HAS to = 0? If it's 0 then lim is 1/2
     
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