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Prove: using $\epsilon$-$\delta$ definition, the limit of both $f$ and $g$ as $(x,y)\to (0,0)$ is $0$.

  1. $f(x,y)=xy$

  2. $g(x,y)=\frac{xy}{x^2 +y^2+1}$

Also, for Q2 can I convert $g(x,y)$ to $m(x,y)/n(x,y)=g(x,y)$ using arithmetic of limits, then prove using $\epsilon$-$\delta$ definition the limit of function $m$ and $n$ separately; then combine the two?

Thanks :)

I wonder if this is correct: $|xy-0|<\epsilon$ given $|x-0|< \delta $ and $|y-0|< \delta $

$|xy-0|< |x-0||y-0|<\delta^2=\epsilon$

therefore: $\delta<\epsilon^{1/2}$

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eddited g(x,y).... – jake Oct 16 '12 at 13:02
why do you care about delta/epsilon? for function of two variable you can do by another way,for first just choose two path ,one for x axis and second on y axis,in both case,if you go on X axis path y is equal to zero,on y axis the same x is equal to zero,so result in both case is zero – dato datuashvili Oct 16 '12 at 13:05
i dont care about epsilon delta :(. my homework question does >< – jake Oct 16 '12 at 13:08
Presumably he cares about delta/epsilon because it is a problem in a chapter on delta/epsilon proofs, and he is meant to apply it. @dato – Thomas Andrews Oct 16 '12 at 13:09
thanks dato! seems i should Google more often :P – jake Oct 16 '12 at 13:28
  1. First notice that $$ |x|=\sqrt{x^2}\le \sqrt{x^2+y^2}=\|(x,y)\|_2,\ |y|=\sqrt{y^2}\le \sqrt{x^2+y^2}=\|(x,y)\|_2 \quad \forall (x,y) \in \mathbb{R}^2. $$ Given $\varepsilon>0$, let $\delta=\sqrt{\varepsilon}$. We have $$ \|(x,y)\|_2\le \delta \Longrightarrow |f(x,y)|=|x||y|\le \|(x,y)\|_2^2 \le \delta^2=\varepsilon $$
  2. Observe that $$ |g(x,y)|=\frac{|f(x,y)|}{x^2+y^2+1}\le |f(x,y)| \quad \forall (x,y) \in \mathbb{R}^2 $$ and use 1.
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What does the subscript "2" mean in the norm notation? – mavavilj Sep 23 '15 at 18:41
It stands for the euclidean norm (as there are many norms on $\mathbb{R}^2$, for instance $\|(x,y)\|_p=(|x|^p+|y|^p)^{1/p}$, where $p$ is a real number with $p\ge 1$). – Mercy King Sep 23 '15 at 22:58

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