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Multivariable limit evaluation

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data


    lim
    (x,y) --> (0,0) [sin(2x) -2x + y]/[x^3+y]

    2. Relevant equations



    3. The attempt at a solution

    so I tried approaching from the x axis and I got

    y=0 lim x-> 0 [sin(2x) -x + 0]/[x^3 + 0] which gives me 0/0

    and from the y axis

    y=0 lim y ->0 which gives me lim y-> 0 y/y = 0/0

    if I try approaching from a line such as y = x I also get 0/0

    what do I do now? L'hopitals rule? If so I'm not sure how to use it here...
     
  2. jcsd
  3. Oct 9, 2011 #2
    I think you'll find that, for any line you choose to approach the origin with, you'll get a single variable limit of 0/0 form. You may apply L'Hospital's Rule for this. Remember that L'Hospital's Rule only works for single variable limits.

    If you are having difficulty getting differing limits, try approaching the origin via the line,

    [tex]y=mx[/tex]

    It'll follow quite naturally from this.
     
  4. Oct 9, 2011 #3
    okay so then using l hopitals rule for this, since it only applies for single variable limits, if I choose y=0 and evaluate it at

    lim x-> 0 I get sin(2x) - 2x/x^3 then using l hopitals rule which is

    lim x->a f'(x)/g'(x) I get

    2 cos(2x) -2/3x^2, evaluating that at x=0 I still get 0/0...

    so what now?
     
  5. Oct 9, 2011 #4

    SammyS

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    For (2 cos(2x) -2)/3x2 , use L'Hôpital's rule again, then again.
     
  6. Oct 9, 2011 #5

    SammyS

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    Try approaching (0,0) along the curve, y = m(x3)
     
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