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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
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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
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thanks dato! seems i should Google more often :P –  jake Oct 16 '12 at 13:28
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1 Answer

  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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