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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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    thanks dato! seems i should Google more often :P2012-10-16

1 Answers 1

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  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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    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 $p\ge 1$).2015-09-23