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Homework Help: Multivariable limit

  1. Sep 28, 2009 #1
    I'm taking multi-variable after having a while off from school, so forgive me if these are simple ones that I just don't "see"

    1. The problem statement, all variables and given/known data
    lim (x, y) --> 0, 0 [tex]\frac{x^2 y^2 e^y}{x^4+4y^2}[/tex]


    lim (x, y) --> (1, 1) [tex]\frac{x-y}{x^3-y}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    The bottom one I feel doesn't exist because as x->0+, the ^3 is making it larger, where as when x->0-, the ^3 is making it smaller. I know this is poor logic; it's just a gut feeling about it. substituting y=mx or similar didn't get me anywhere. Plugging obviously doesn't work. I don't see anyway to simplify, but maybe there is a way. I also don't think polar coordinates will work for either.

    I really think I only need a hint to the method of solution, so don't go solving the whole thing for me.
    Last edited: Sep 28, 2009
  2. jcsd
  3. Sep 28, 2009 #2


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    For the bottom one, suppose you approach along the path y=x^3? For the top one you don't have to worry about the factor e^y, why not? Can you handle the rest?
  4. Sep 28, 2009 #3
    For the top one, e^y -->1 as y->0. I wasn't sure if saying that and then doing the rest of the limit was a legal move. Should be easy now (I see one way by polar coordinates and then squeeze theorem, I think)

    For the bottom one, I tried that, but doesn't that just 'break' the function--as in, we can't tell what it will be by that method? The way I thought the "two different limits for two different paths" property worked was that it would only work if you found two actual different paths. Taking y=x^3 makes one path not exist; is that enough to make the limit of the function not exist?
  5. Sep 28, 2009 #4


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    If the function has a limit, it has to approach that limit along all paths. If the function doesn't even exist along a path, then the limit doesn't exist. If you think about what's happening 'close' to the path y=x^3 then you'll see the function becomes large without bound.
  6. Sep 28, 2009 #5
    ha, I had already went back and put it on paper, and as soon as I did the behavior 'close' to y=x^3 became clear. thanks! hopefully my brain will uncrustify itself soon...
  7. Sep 30, 2009 #6
    follow up on the bottom one: Subbing y=x^3 will not work because it is not within the definition of a limit. However, comparing the substitutions y=x and y=1 does work to prove that it indeed DNE.
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