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Limit with 2 variables

  1. May 12, 2012 #1
    Hello, I've been trying to solve the following limit for the last hour or so. I'm pretty much at a loss as to how to solve it (I need to do a lot of these kinds of problems, but I need to understand how to approach them first), so any pointers in the right direction would be appreciated.

    [tex]\lim_{\substack{x\rightarrow -1\\y\rightarrow 1}} \frac{x^2 - y^2}{xy + y^2 + x + y}[/tex]

    Thankyou.
     
  2. jcsd
  3. May 12, 2012 #2
    I believe you need to learn multivariable calculus basics? For limits,

    http://tutorial.math.lamar.edu/Classes/CalcIII/Limits.aspx

    It is a rather good webiste to learn Calculus basics in general, so I recommend that you explore it, as for your needs. Remember to try out the easier examples to get a good hang of it!
     
  4. May 12, 2012 #3
    Thanks for the reply, but for this problem I don't think that site will help me very much, as I don't have a problem tackling any of the four problems it poses (the first two are just simple substitutions, and the second two just require the Two Path Test to demonstrate that no limit is approached). This problem differs from all those examples and I'm not how to solve it.
     
  5. May 12, 2012 #4
    I guess I misunderstood that you needed to learn multivariable calculus from the first post..

    Anyway, try reducing that expression to something simpler. (Hint: factorize the numerator and denominator, and see if there's something you can eliminate so that the problem becomes a simple substitution sum)
     
  6. May 12, 2012 #5
    I've tried it, I can factorise the numerator (x+y)(x-y), but I can't factorise the demoninator, and don't see any available substitutions
     
  7. May 12, 2012 #6
    Why not? The terms are laid out to make it rather easy to factorize :smile:

    You can write the denominator as...

    [tex]y(x+y) + (x + y)[/tex]

    Can you take it from here?
     
    Last edited: May 12, 2012
  8. May 12, 2012 #7
    Yes I can, thankyou very much for your help!
     
  9. May 12, 2012 #8
    You're welcome! :smile:
     
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