# Squeeze Theorem - Multivarible question

by pbxed
Tags: multivarible, squeeze, theorem
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 P: 14 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)
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,568 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.
 P: 14 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.
 P: 14 Squeeze Theorem - Multivarible question 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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