Squeeze Theorem - Multivarible question

  1. Hi,

    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. HallsofIvy

    HallsofIvy 40,310
    Staff Emeritus
    Science Advisor

    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. 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. 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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