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2 multivariable limit questions.

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Hello, I need help on the following questions, mainly to make sure i did them right.

    for both, lim (x,y) -> (0,0)

    1) [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP3619hd2e84h398acc80000110c0eac25d415ha?MSPStoreType=image/gif&s=57&w=72&h=44 [Broken]

    2) [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP43819hd2e19gfb031c4000051d54bgf834ga83b?MSPStoreType=image/gif&s=47&w=80&h=40 [Broken]


    3. The attempt at a solution

    1) so for this one I multiplied by the conjugate of the denominator.
    So what i end up with is as follows

    x(x-y)(sqrt x + sqrt y) / (x-y)

    x-y cancels so I am left with

    x(sqrt x + sqrt y)

    so I get the limit as 0 when i plug in x and y.

    2)
    This one I am clueless about what to do. I keep getting an indeterminate form 0/0. I tried L'hopital's rule but it doesn't really help. Any ideas? Do i need to use some trigonometric identity here?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 10, 2011 #2

    SammyS

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    (1) This is correct.

    (2) Try approaching (0, 0) along some line which passes through the origin.

    Alternatively, you can see what limit you get if you approach the x-axis along the line x = b, for some constant, b ≠ 0. Then, if that limit exists, approach (0,0) with that expression. Do similar for the y-axis.
     
  4. Oct 10, 2011 #3

    SammyS

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    More on (1).

    Of course you can only approach (0,0) from the first quadrant or the positive x or y axis.
     
  5. Oct 10, 2011 #4
    Thanks for the reply.

    I'm still a bit lost on #2. Using the first method such as taking a curve, say y=x that passes through the origin i still end up with 0/0, i tried L'hopitals rule after using y=x for example and I still don't get anywhere.

    as for taking some x=b, so suppose b=1, then I would lim x,y -> 1,0. which also gives me an indeterminate form 0/0?
     
  6. Oct 10, 2011 #5

    SammyS

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    I don't see the image of the problem any longer.
     
  7. Oct 10, 2011 #6
    Sorry bout that. This is the question.

    lim
    x,y -> 0,0

    [cos(xy)-1]/[(x^2)(y^2)]
     
  8. Oct 10, 2011 #7

    SammyS

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    Suppose y=mx.

    [itex]\displaystyle \frac{\cos(xy)-1}{x^2\,y^2}\to \frac{\cos(m\,x^2)-1}{m^2x^4}[/itex]

    Now for L'Hôpital
     
  9. Oct 10, 2011 #8
    Hmm looks like there is an error somewhere in the latex code, can't see what is written, but I'll try subbing in mx like you said and see what I can do from there with L'hopitals. I might have done something wrong then.
     
  10. Oct 10, 2011 #9

    SammyS

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    Apply L'Hôpital twice for these.
     
  11. Oct 10, 2011 #10
    Using L'hopitals will make me get rid of the x from the denominator eventually but i think it will take more than 2 times. This is what i have so far.

    using L'hopitals the first time I got:

    -2mx sin(mx^2) /( 4x^3)(m^2)

    which is still 0/0 as x -> 0

    second time:

    [-2m sin(mx^2) - 4m^2x^2 cos(mx^2)]/](12x^2)(m^2)]

    which is still 0/0

    unless i did something wrong, i can keep using l'hopitals to remove x from the denominator and should not give me 0/0 but it will be a total of 4 times I have to use it.
     
  12. Oct 10, 2011 #11

    SammyS

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    Cancel what you can before proceeding.
     
  13. Oct 10, 2011 #12
    ...right! stupid mistakes :(

    thanks so much for the help!
     
  14. Oct 10, 2011 #13

    SammyS

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    I get a similar result for this method.
     
  15. Oct 10, 2011 #14

    SammyS

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    Now for the bad news.

    This all looks great, but it doesn't prove the limit is -1/2 or whatever we got.

    If any were different than -1/2, then the limit DNE.

    This may be good enough for your course at this time.
     
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