Squeeze Theorem - Multivarible question

  1. Aug 7, 2011 #1

    I'm having a lot of difficulty with finding limits of multivariable functions. A question like this comes up every year in the final exam and it will always ask for use of the squeezing theorem.

    1. The problem statement, all variables and given/known data

    (a) Suppose that
    f(x, y) = 1 +(5x2y3)/x2 + y2

    for (x, y) =/= (0, 0)

    and that f(0, 0) = 0. By applying the Squeezing Rule to |f(x, y) − 1|, or otherwise, prove

    that f(x, y) -> 1 as (x, y) -> (0, 0).

    2. Relevant equations

    3. The attempt at a solution

    I understand that in order for a limit to exist that no matter what direction we approach (0,0) we must compute the same value. From x-axis and y-axis it seems that the limit is indeed 1. I also get the intuition of squeeze theorem that

    lim (x,y) -> (a,b) g(x) <= lim (x,y) -> (a,b) f(x) <= lim (x,y) -> (a,b) h(x)

    so lim g(x) = lim h(x) then we have found our lim f(x)

    What I'm really confused about is how we set up the squeeze inequality that I see in some textbooks.

    Would it be something like this ?

    1 =< (5x2y3)/(x2 + y2) <= (I have no idea how you would find an expression on the RHS)
  2. jcsd
  3. Aug 7, 2011 #2


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    It's almost impossible to tell what you mean. If you don't want to use LaTeX or html tags, at least use "^" to indicate a power. I think your "5x2y3" is supposed to mean 5x^2y^3. But you also need parentheses. Do you mean
    1)1 +((5x^2y^3)/x^2) + y^2
    2) (1+ 5x^2y^3)/ (x^2+ y^2)
    3) 1+ 5x^2y^3/(x^2+ y^2)?

    I suspect you mean the third but I cannot be certain

    I recommend casting into polar coordinates, then getting you "squeeze" by observing that sine and cosine are always between -1 and 1.
  4. Aug 7, 2011 #3
    Oh, sorry I did mean the third one. I copied and pasted directly from a pdf file and it messed up the formatting without me realising sorry. Ill try the polar coordinates now thx.
  5. Aug 7, 2011 #4
    Okay, So I subbed in x = r cos (alpha) and y = r sin (alpha) and simplified the expression down to

    1 + 5r^3(cos(alpha))^2(sin(alpha))^3

    My question is, because as the lim r-> 0 then isn't the limit just 1 (which is what I wanted to show) and I wouldn't have to use the squeeze theorem if doing this question in polar coordinates ?
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