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CALC III Finding limits for a function of two variables.

  1. Feb 3, 2012 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    Find the limit of f(x,y) = [itex]\frac{x^{2}-y^{2}}{\sqrt{x^{2}+y^{2}+81}-9}[/itex]
    as (x,y) [itex]\rightarrow[/itex] (0,0)

    3. The attempt at a solution

    Ok, I looked at the examples in the book, and it seems pretty straight forward... I first look at f(x,0) to see what happens as we approach along the x-axis. The equation becomes 0/0, which is indeterminate. I did the same thing for f(0,y) to see what happens as we approach along the y-axis, which is also indeterminate. This would mean that the limit doesn't exist, according to the book. I even also approached on a different line, y=x, but that is also indeterminate.

    That is the wrong answer however, according to the software. I thought maybe I would have to use l'hopital's rule for limits, but I'm not sure if I am supposed to do that, or if I would need to do it twice (with respect to x AND y). I started to take the partial with respect to x and it got very messy, so I wasn't sure if that was what I needed to do.

    This is the very first question we are assigned on limits in calc 3.

    EDIT:

    I also just tried setting [itex]\sqrt{x^{2}+y^{2}+81}\neq9[/itex] and I got that y cannot be equal to plus or minus x, but I'm not really sure what to do with that.
     
  2. jcsd
  3. Feb 3, 2012 #2

    Dick

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    Try multiplying numerator and denominator by sqrt(x^2+y^2+81)+9 and simplify the denominator. Then try your different paths approaching 0,0.
     
  4. Feb 3, 2012 #3

    ElijahRockers

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    Oya.

    Alright, I did that. It simplifies to [itex]\frac{(x^{2}-y^{2})(\sqrt{x^{2}+y^{2}+81}+9)}{x^2+y^2}[/itex]

    I tried approaching along the x axis and got 18.
    When I approach along the y axis, however, I get -18, because the first y2 is negative.

    This would mean the limit doesn't exist, which it does, so obviously my algebra is wrong somewhere. Any idea where I'm going wrong here...?
     
  5. Feb 3, 2012 #4

    jbunniii

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    Why are you so sure the limit exists? I see nothing wrong with your calculations, and they show that the limit does NOT exist.
     
  6. Feb 3, 2012 #5

    SammyS

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    Why do you say the limit does exist ?
     
  7. Feb 3, 2012 #6

    ElijahRockers

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    Because I answered the question that the limit does not exist before I posted here, and it told me it was wrong. :P

    But I guess even the coursework software can be wrong?
     
  8. Feb 3, 2012 #7

    jbunniii

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    I think your coursework software is wrong, assuming the problem is exactly as you stated it here.
     
  9. Feb 3, 2012 #8

    ElijahRockers

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    I think I transcribed the question correctly, but just for confirmation, I took a screenshot.

    NuNbd.png
     
  10. Feb 3, 2012 #9

    ElijahRockers

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    OHHH. Duh. I see it. I guess I just had to take a screenshot to notice it. Thank you everybody.

    </oblivious>

    But while I have everyone's attention, I would like some clarification on something.

    So I found two paths that agree on a limit, but there are literally an infinite number of paths to approach the point from. How can I be sure every single path agrees on the limit? The way I understand it, even if a single path doesn't agree, then the limit doesn't exist.

    How can I be sure that 18 is the correct answer? (it is, according to the software)
     
  11. Feb 3, 2012 #10

    jbunniii

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    Good question! Why don't you try a similar manipulation with the new function - clear the denominator and simplify. What do you end up with?
     
  12. Feb 3, 2012 #11

    ElijahRockers

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    18.

    Thanks very much... I am a little rusty, but after a few stupid mistakes, I am slowly getting back in the groove.
     
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