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pbxed
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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.
(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).
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)
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.
Homework Statement
(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).
Homework Equations
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)