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As the topic, how to prove by $\epsilon$-$\delta $ approach $\lim_{(x,y)\rightarrow (0,0)}\frac {x^n-y^n}{|x|+|y|}$ exists for $n\in \mathbb{N}$ and $n>1$

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    @Mathematics, I think it should be interesting you verify why the result doesn't holds for $n=1$.2012-03-05

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Let $\epsilon>0$. There is an $\delta>0$ such that $\xi^n\leq \epsilon\xi$ for all $\xi\in[0,\delta)$. Then if $(x,y)\in(-\delta,\delta)\times(-\delta,\delta)$ and $(x,y)\neq0$, we have $\left|\frac{x^n-y^n}{|x|+|y|}\right|\leq\frac{|x|^n+|y|^n}{|x|+|y|}\leq\frac{\epsilon(|x|+|y|)}{|x|+|y|}=\epsilon.$ We win.

P.S. Mathematics wants to know how to prove the existence of $\delta$. One can proceed like Neal suggests, or various variations of that idea. A simpler approach is the following. If $0\leq\xi\leq\min\{\epsilon,1\}$ then $0\leq\xi^n=\xi\cdot\xi\cdot\xi^{n-2}\leq\epsilon\cdot\xi\cdot1=\epsilon\xi$ because $0\leq\xi^{n-2}\leq1$. This means that we can take $\delta=\min\{\epsilon,1\}$. :)

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    I *am* curious about the reason for the downvotes :)2012-03-05
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Hint 2: Working from what you have (as per your comment on user22705's answer), observe that $\begin{eqnarray}(|x|+|y|)^2&=&x^2+y^2+2|x||y|\\ &\leq& x^2+y^2+2x^2+2y^2\\ &=&3(x^2+y^2)\end{eqnarray}$

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A HINT is to rewrite the numerator using the following identity: $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+….+b^{n-1})$ then use the triangle inequatlity.

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    Mathematics: Coming back to your first comment: surely you can compare $|x|$ and $(x^2+y^2)^{1/2}$! And $|y|$ and $(x^2+y^2)^{1/2}$. Thus bounding $|x|+|y|$ by a multiple of $(x^2+y^2)^{1/2}$ should be doable, no?2012-03-04
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You may use that

$\left|\frac{x^n-y^n}{|x|+|y|}\right|\leq \frac{|x|^n-|y|^n}{|x|+|y|}\leq \frac{|x|}{|x|+|y|}|x|^{n-1}+\frac{|y|}{|x|+|y|}|y|^{n-1}\leq|x|^{n-1}+|y|^{n-1}.$

Since you impose $x^2+y^2< \delta \leq 1$ you have $|x|, |y|<1\Rightarrow |x|^{n-1}<|x|,\ |y|^{n-1}<|y|.$

Then you have

$|x|^{n-1}+|y|^{n-1}<|x|+|y|\leq 2\sqrt{x^2+y^2}$.

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    The definition of limit will be more important than this naive example: We say that $lim_{(x,y)\to (a,b)}f(x,y)=L$ when for all \epsilon >0 there is a \delta=\delta_{\epsilon}>0 such that ||(x,y)-(a,b)||<\delta \Rightarrow ||f(x,y)-L||<\epsilon2012-03-05