Easy delta/epsilon proof of a multivariable limit

[pureza]

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:

\lim_{(x,y) \to (1,1)}\frac{xy}{x+y}

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

Any hints?

Thank you!

 

[HallsofIvy]

Well, what have you done? I assume you want to prove that

"Given \epsilon> 0 there exist \delta> 0 such that if \sqrt{(x-1)^2+ (y-1)^2}<\delta then
\sqrt{\frac{xy}{x+y}- 1}<\epsilon"

 

[pureza]

Actually I think I might have solved it:

I want to prove that

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

Now,

\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| by the triangle inequality.

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

This way,

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

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

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

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

What's wrong with my rationale? :-)

 

[jgens]

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

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

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 ≤ ε2^1/2. 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 ≤ ε2^1/2 but ε < |x-1|+|y-1|.

 

[pureza]

Obviously, you are right (thank you). Take ε=1. I can choose a point in the circle with radius \sqrt{2} that is not inside the square |x-1|+|y-1|&lt;1.

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 δ\leq ε, then I know that any point inside the circle with radius δ verifies |x-1|+|y-1|&lt;ε, which, if the first part of my derivation is correct, implies that

\left|\frac{xy}{x+y}- \frac{1}{2}\right|&lt;ε

Is this correct?