proving by $\epsilon$-$\delta $ approach that $\lim_{(x,y)\rightarrow (0,0)}\frac {x^n-y^n}{|x|+|y|}$exists  for $n\in \mathbb{N}$ and $n>1$ 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 $n>1$ 
 A: 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 $\delta=\min\{\epsilon,1\}$. :)
A: Hint 2: Working from what you have (as per your comment on user22705's answer), observe that $$\begin{eqnarray}(|x|+|y|)^2&=&x^2+y^2+2|x||y|\\
&\leq& x^2+y^2+2x^2+2y^2\\
&=&3(x^2+y^2)\end{eqnarray}$$
A: 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.
A: 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
$|x|^{n-1}+|y|^{n-1}<|x|+|y|\leq 2\sqrt{x^2+y^2}$.
