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Limit of function of two variables using Squeeze Theorem?

  1. Feb 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine the set of points at which the function is continuous:

    f(x,y)= (x^2)(y^3)/[2x^2 + y^2] if (x,y) is not equal to (0,0)
    = 1 if (x,y) is equal to (0,0)

    2. Relevant equations

    Definition of the limit of a function of two variables:

    lim f(x,y) as (x,y)-->(a,b)=L if for every number E>0 there is a corresponding number D>0 such that if 0<sqrt[(x-a)^2 + (y-b)^2]< D then the absolute value of f(x,y) - L < E.

    3. The attempt at a solution
    f is continuous on R^2 except at possibly the origin. Since x^2< or = 2x^2 + y^2, the absolute value of (x^2)(y^3)/[2x^2 + y^2] is < or = the absolute value of y^3 (which has a limit of 0 as (x,y) approaches 0). Now how can I use this to determine the limit of f(x,y)? I'm guessing it has something to do with the Squeeze Theorem, or a direct application of the definition of the limit of a function of two variables.
     
  2. jcsd
  3. Mar 1, 2009 #2
    Compute f(x,0) for different values of x.
     
  4. Mar 1, 2009 #3

    HallsofIvy

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    Or f(0,y) for different values of y! :rofl: That is, find [itex]\lim_{x\rightarrow 0} f(x,0)[/itex] or [itex]\lim_{y\rightarrow 0} f(0,y)[/itex]

    In order for the function to be continuous at (0,0), the limit must exist and it must be equal to the value of f(0,0).
     
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