2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $$\left|{4x^2y^3+x^4y-y^5\over(x^2+y^2)^2}\right|\leq6 r<6\delta=\epsilon\ .$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.