# Easy delta/epsilon proof of a multivariable limit

by pureza
Tags: limit definition
 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: $\lim_{(x,y) \to (1,1)}\frac{xy}{x+y}$ Obviously, the result is $1/2$, but I'm unable to prove it! Any hints? Thank you!
 PF Patron Sci Advisor Thanks Emeritus P: 38,424 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 $$\sqrt{\frac{xy}{x+y}- 1}<\epsilon$$"
 P: 3 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? :-)
P: 1,391

## Easy delta/epsilon proof of a multivariable limit

 Quote by pureza 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 ≤ 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|.
 P: 3 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 $|x-1|+|y-1|<1$. 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 $|x-1|+|y-1|<ε$, which, if the first part of my derivation is correct, implies that $$\left|\frac{xy}{x+y}- \frac{1}{2}\right|<ε$$ Is this correct?

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