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LIMITS - Multi-variable calculus

  1. Oct 6, 2004 #1
    Hi,

    I have an example from my prof. He said, in evaluating limits, we can use L1 (along x-axis), L2 (along y-axis), and L3 (y = kx). After that, if the limits come to the same finite number, then the limit might exist. Here's the example.

    [tex]\lim_{\substack{x\rightarrow 0\\y\rightarrow 0}} f(x,y)=\frac{x^3y}{x^6+y^2}[/tex]

    The L1, L2, L3 limits all come to zero.

    Then being the prof because he is smart, he used L4 = y = x^3, and showed the limit equals some other number, which proved the limit doesn't exist.

    [tex]\lim_{\substack{x\rightarrow 0\\y\rightarrow x^3}} f(x,y)=\frac{x^3y}{x^6+y^2}=1/2[/tex]

    How can I, as a student, come up with some "random" function such as [itex]y=x^3[/itex] to show that the limit does not exist??

    How do I know if the limit exist for sure?? Is there a way to tell??

    :confused: :confused: :confused: Thanks for the help! :smile:
     
    Last edited: Oct 6, 2004
  2. jcsd
  3. Oct 6, 2004 #2

    Galileo

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    That's funny. It seems to me the limit does exist and equals 0.

    The easiest way to see this is to notice the function is continuous. Then you can just plug in x=y=0.

    How on earth did your prof showed
    [tex]\lim_{\substack{x\rightarrow 0\\y\rightarrow x^3}} f(x,y)=\frac{xy}{3+x^2y^2}=1/2[/tex]??
     
  4. Oct 6, 2004 #3

    arildno

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    Your professor is dead wrong.
    1) xy is continuous (easily proven)
    2) [tex]3+x^{2}y^{2}[/tex] is continuous, and moreover, greater than zero (easily proven)
    3) This means that the fraction is also continuous (at all points), and hence, has limits there as well
     
  5. Oct 6, 2004 #4
    Oh crap. I posted the wrong limit! :blushing: :uhh:
    My apologies!

    [tex]\lim_{\substack{x\rightarrow 0\\y\rightarrow 0}} f(x,y)=\frac{x^3y}{x^6+y^2} = 1/2[/tex]

    Sorry again for the carelessness! :blushing: (extended apologoes to Galileo and Arildno)
     
    Last edited: Oct 6, 2004
  6. Oct 6, 2004 #5

    arildno

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    Your professor was right after all, then..:wink:
     
  7. Oct 6, 2004 #6
    Yeah....so how can I know how to plug a function such as y=x^3 to show that the limit DNE??

    He only taught us to use the L1, L2, L3 method.
     
  8. Oct 7, 2004 #7

    Galileo

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    There isn't a surefire way to solve a limit question (or any problem for that matter).
    In this case you just have to notice that the expression y=x^3 will make the two terms in the denominator both of degree 6.
     
  9. Oct 7, 2004 #8

    HallsofIvy

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    There is no way of knowing WHAT "counter-example" will work- no matter how many curves you use on which the limit is the same, you can't be sure there isn't some curve on which it is different.

    What you COULD do is convert to polar coordinates. That way, the distance to (0,0), which is what is important, is just the single variable r.

    In this particular case, You would find that the limit also depends on θ so taking r-> 0 does not give a limit.
     
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