Epsilon Delta Proof? I always have trouble with understanding the intuition/process of $\epsilon$-$\delta$ proofs.  Could anyone assist me with understanding the solution to the following:
Show that $f$ is continuous at $(0,0)$.
$$f(x,y) =\begin{cases} \dfrac{x^4}{x^2+y^2}&\text{ for }(x,y)\neq (0,0)\\0&\text{ for }(x,y)  = (0,0)\end{cases}$$ 

The proposition under consideration is: For every $\epsilon > 0,$ a $\delta > 0$ exists such that if $|x|, |y| < \delta$ then $|x^{4}/(x^{2} + y^{2})| < \epsilon.$
Let $\epsilon > 0.$ If $|x| = |y|$, then 
$$|\frac{x^{4}}{x^{2} + y^{2}}| = \frac{x^{2}}{2} < \frac{\delta^{2}}{2},$$
so, for $|x^{4}/(x^{2}+y^{2})|$ to be $< \epsilon$ it suffices to take
$$\delta := \sqrt{2\epsilon}.$$
Without loss of generality, let $|x| < |y|$ (because the case where $|x| > |y|$ leads to the same result.). Then again we have
$$|\frac{x^{4}}{x^{2} + y^{2}}| < \frac{x^{4}}{2x^{2}} = \frac{x^{2}}{2} < \frac{\delta^{2}}{2},$$
so taking
$$\delta := \sqrt{2\epsilon}$$
suffices.
Here we have completed the proof.
