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Easy delta/epsilon proof of a multivariable limit

by pureza
Tags: limit definition
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pureza
#1
Jan18-12, 11:28 PM
P: 3
Hi,

I'm trying to wrap my head around epsilon/delta proofs for multivariable limits and it turns out I became stuck on an easy one!

The limit is:

[itex]\lim_{(x,y) \to (1,1)}\frac{xy}{x+y}[/itex]

Obviously, the result is [itex]1/2[/itex], but I'm unable to prove it!

Any hints?

Thank you!
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#2
Jan19-12, 09:02 AM
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Well, what have you done? I assume you want to prove that

"Given [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]\sqrt{(x-1)^2+ (y-1)^2}<\delta[/itex] then
[tex]\sqrt{\frac{xy}{x+y}- 1}<\epsilon[/tex]"
pureza
#3
Jan19-12, 09:48 AM
P: 3
Actually I think I might have solved it:

I want to prove that

Given [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]\sqrt{(x-1)^2+ (y-1)^2}<\delta[/itex] then
[tex]\left|\frac{xy}{x+y}- \frac{1}{2}\right|<\epsilon[/tex]

Now,

[tex]\left|\frac{xy}{x+y}- \frac{1}{2}\right|=\left|\frac{2xy-(x+y)}{2(x+y)}\right|=\left|\frac{xy + xy-x-y)}{2(x+y)}\right|=\left|\frac{x(y-1)+y(x-1)}{2(x+y)}\right|\leq\left|\frac{x(y-1)}{2(x+y)}\right|+\left|\frac{y(x-1)}{2(x+y)}\right|[/tex] by the triangle inequality.

Now, I know that since [itex]{(x,y) \to (1,1)}[/itex] I can make [itex]x[/itex] and [itex]y[/itex] positive so that [itex]\left|\frac{x}{x+y}\right|\leq1[/itex] and [itex]\left|\frac{y}{x+y}\right|\leq1[/itex]. [I'm no sure about this argument]

This way,

[tex]\left|\frac{x(y-1)}{2(x+y)}\right|+\left|\frac{y(x-1)}{2(x+y)}\right|\leq|y-1|+|x-1|[/tex]

But this is just a square with side 2 around the point [itex](x, y)[/itex]. This square is entirely contained by the circle with center [itex](x, y)[/itex] and radius [itex]\sqrt{2}[/itex].

So, when given some [itex]ε>0[/itex] if I set [itex]δ=min(1, ε\sqrt{2})[/itex] and I choose a point [itex](x, y)[/itex] such that [itex]\sqrt{(x-1)^2+ (y-1)^2}<ε\sqrt{2}[/itex], then certainly [itex]|y-1|+|x-1|<ε[/itex] and, by extension,

[tex]\left|\frac{xy}{x+y}- \frac{1}{2}\right|<ε[/tex]

What's wrong with my rationale? :-)

jgens
#4
Jan19-12, 11:43 AM
P: 1,622
Easy delta/epsilon proof of a multivariable limit

Quote Quote by pureza View Post
So, when given some [itex]ε>0[/itex] if I set [itex]δ=min(1, ε\sqrt{2})[/itex] and I choose a point [itex](x, y)[/itex] such that [itex]\sqrt{(x-1)^2+ (y-1)^2}<ε\sqrt{2}[/itex], then certainly [itex]|y-1|+|x-1|<ε[/itex] and, by extension,

[tex]\left|\frac{xy}{x+y}- \frac{1}{2}\right|<ε[/tex]
I did not read anything other than this line so I cannot comment on anything else you wrote, but this is wrong. Take (x,y) such that ε < [(x-1)2+(y-1)2]1/2 ≤ ε21/2. Then notice that [(x-1)2+(y-1)2]1/2 ≤ |x-1|+|y-1| to find (x,y) with [(x-1)2+(y-1)2]1/2 ≤ ε21/2 but ε < |x-1|+|y-1|.
pureza
#5
Jan19-12, 12:13 PM
P: 3
Obviously, you are right (thank you). Take [itex]ε=1[/itex]. I can choose a point in the circle with radius [itex]\sqrt{2}[/itex] that is not inside the square [itex]|x-1|+|y-1|<1[/itex].

How about this: if I choose a point inside the circle with radius ε, then that point is also inside the square with side 2ε, because the circle is contained within the square.

Therefore, if I set [itex]δ\leq ε[/itex], then I know that any point inside the circle with radius δ verifies [itex]|x-1|+|y-1|<ε[/itex], which, if the first part of my derivation is correct, implies that

[tex]\left|\frac{xy}{x+y}- \frac{1}{2}\right|<ε[/tex]

Is this correct?


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