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$\Bbb R^2$ is using the Euclidean metric, $\Bbb R$ is using the standard $|y-x|$ metric.

We define $f:\Bbb R^2\rightarrow\Bbb R$ by

$$f(x,y) = \left\{\begin{array}{ll} \frac{x^6+y^6}{x^2+y^2} & : (x,y)\neq(0,0)\\ 0 & : (x,y)=(0,0) \end{array} \right.$$

I need to show this is continuous at $(0,0)$. While this is easy for me to understand in a more intuitive way, since $\frac{x^6+y^6}{x^2+y^2}$ would clearly have a discontinuity at $(0,0)$, I'm rubbish at proving these things using $\epsilon$ and $\delta$.

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    $\begingroup$ Hint: $x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)$ $\endgroup$ – David Dec 4 '14 at 4:12
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Take $x=rcos\theta,y=rsin\theta $

then $r^2=x^2+y^2$

So $|\frac{x^6+y^6}{x^2+y^2}|=|r^4(cos^6\theta+sin^6\theta)|\leq 2r^4=2(x^2+y^2)^2<\epsilon$

whenever $x^2<\frac{\epsilon}{4}$ and $y^2<\frac{\epsilon}{4}$

Taking $\delta =\sqrt\frac{\epsilon}{4}$ we get the required answer

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$x^6+y^6 \le (|x|^3+|y|^3)^2$.

Cauchy Schwarz gives $|x|^3+|y|^3 = \langle (x^2,y^2), (|x|, |y|)\rangle \le \sqrt{x^4+y^4}\sqrt{x^2+y^2}$, combining gives

$x^6+y^6 \le (x^4+y^4)(x^2+y^2)$.

If $0<x^2+y^2 \le 1$, then $x^4+y^4 \le x^2+y^2$, from which we get ${x^6+y^6 \over x^2+y^2} \le x^2+y^2$, from which the result follows.

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