Suppose $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is defined as $$(x,y) \longmapsto \left\{ \begin{array}{cl} \dfrac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2} & \mbox {if } (x,y) \neq (0,0) \\ \\ 0 & \mbox {if } (x,y) = (0,0) \end{array} \right. $$

I have shown that $f$ is continuous at $(0,0)$ using polar coordinates, but I am really trying to improve my $\epsilon-\delta$ proofs for such limits. I always end up getting confused with the inequalities. (Sorry if the question is a bit repetitive here)

So for $\epsilon > 0$ I need to find a $\delta$ such that $\|(x,y)\| =\sqrt{x^2 + y^2} < \delta$ implies $|f(x,y)| < \epsilon$

I have tried to work backwards using the inequality $(x^2+y^2)^2 \geq 4x^2y^2$ $$\begin{align*} \left|\frac{4x^2y^3 +x^4y - y^5}{(x^2+y^2)^2}\right|&=\left|\frac{4x^2y^3}{(x^2+y^2)^2} + \frac{x^4 y}{(x^2+y^2)^2} - \frac{y^5}{(x^2+y^2)^2}\right|\\ &\leq\left|\frac{4x^2y^3}{(x^2+y^2)^2}\right| + \left|\frac{x^4 y}{(x^2+y^2)^2}\right| + \left|\frac{y^5}{(x^2+y^2)^2}\right|\\ &\leq\left|\frac{4x^2y^3}{4x^2y^2}\right| + \left|\frac{x^4 y}{4x^2y^2}\right| + \left|\frac{y^5}{4x^2y^2}\right|\\ &= \left|y\right| + \left|\frac{x^2}{4y}\right| + \left|\frac{y^3}{4x^2}\right| \end{align*}$$

Got stuck here and not sure if my approach is correct.

  • $\begingroup$ Having $x$ and $y$ in the denominator is going to cause you horrible problems with things blowing up. Just use the fact that $|x|$ and $|y|$ are both $\le (x^2+y^2)^{1/2}$. $\endgroup$ – Ted Shifrin Jan 28 at 19:16

The essential point is that $r:=\sqrt{x^2+y^2}$ governs the distance from some point $(x,y)$ to $(0,0)$. In particular $|x|\leq \sqrt{x^2+y^2}=r$, and similarly $|y|\leq r$. It follows that $$\left|{4x^2y^3+x^4y-y^5\over(x^2+y^2)^2}\right|\leq(4+1+1){r^5\over r^4}=6r\ .\tag{1}$$ Given an $\epsilon>0$ choose $\delta:={\epsilon\over6}$. If $|(x,y)|=r<\delta$ then, according to $(1)$, we have $$\left|{4x^2y^3+x^4y-y^5\over(x^2+y^2)^2}\right|\leq6 r<6\delta=\epsilon\ .$$

  • $\begingroup$ Thank a lot for your answer. It made it very simple :) $\endgroup$ – mike Jan 28 at 22:05

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