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Multivariable Calculus proof!

  1. Sep 21, 2011 #1
    I need to show that

    limit (|x|^a*|y|^b) / (|x|^c+|y|^d) = 0
    (x,y)->(0,0)

    when a,b>=0; c,d>0; with a/c + b/d > 1

    Does anyone have some tips for starting off the proof?
     
  2. jcsd
  3. Sep 21, 2011 #2
    Show that if [itex]a, b \geq 0[/itex] and [itex] c, d > 0,[/itex] with [itex]\frac{a}{c} + \frac{b}{d} > 1,[/itex] then:

    [itex]lim_{\vec{x}\rightarrow\vec{0}} \frac{|x|^{a}|y|^{b}}{|x|^{c}+|y|^{d}} = 0[/itex]

    Sorry guys, totally forgot about latex! Here's a more readable version...
     
  4. Sep 21, 2011 #3

    HallsofIvy

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    Do you mean [itex]\lim_{x\to 0}[/itex] or [itex]\lim_{(x,y)\to 0}[/itex]. If the former, it is pretty easy! As x goes to 0, for any y, the numerator goes to 0. If y is not 0, the denominator goes to [itex]|y|^d[/itex] and, of course, [itex]0/|y|^d= 0[/itex]. If y is 0, then the function is 0 for all x so its limit is still 0.

    If the problem is [itex]\lim_{(x,y)\to 0}[/itex],l that's a bit harder.
     
  5. Sep 21, 2011 #4
    It's the second one. I tried to put the vector lines over the x and 0 lol.

    Yeah I'm not sure how to start this! Should I try using Squeeze theorem with something? Or the definition of a limit?
     
  6. Sep 22, 2011 #5
    I figured it out, decided to post the answer just in case someone else has the same kind of question sometime...
    basically just change the numerator to:
    (|x|^c)^(a/c) (|y|^d)^(b/d)
    and use the inequalities:
    |x|^c <= |x|^c + |y|^d
    |y|^d <= |x|^c + |y|^d
    then cancel out and use squeeze theorem
     
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