Epsilon Delta Proof?

Question

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 $(0,0)$.

$$f(x,y) =\begin{cases} \dfrac{x^4}{x^2+y^2}&\text{ for }(x,y)\neq (0,0)\\0&\text{ for }(x,y) = (0,0)\end{cases}$$

Answer 1

$$\left|\frac{x^4}{x^2+y^2}\right|=\left|x^2\cdot\frac{x^2}{x^2+y^2}\right|\leq |x|^2\leq x^2+y^2=||(x,y)||^2$$

So, suppose we are given $\epsilon>0$. We need to give $\delta$ such that if $||(x,y)||<\delta$ then $\left|\frac{x^4}{x^2+y^2}\right|<\epsilon$.

What about $\delta=\sqrt{\epsilon}$?

Answer 2

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.