5
$\begingroup$

I have the limit $$\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4},$$ and would like to show with an $\epsilon-\delta$ proof that it is zero. I know with a situation like $$\left|\frac{x^4y}{x^4+y^4}\right|\leq y$$ or something similar, but I can't find a way to do the same thing here, as no single term in the numerator is of sufficient degree, although I think I could get this with a small hint.

$\endgroup$
  • 2
    $\begingroup$ You can use $2ab \leqslant a^2 + b^2$. $\endgroup$ – Daniel Fischer Oct 1 '13 at 21:48
  • $\begingroup$ Or this is one place a polar coordinates substitution can work really nicely, as $|\cos^3\theta\sin^2\theta|\le 1$ (we could get a better upper bound, but why bother?). $\endgroup$ – Ted Shifrin Oct 1 '13 at 21:51
  • $\begingroup$ Thanks guys, the first hint especially was quite helpful. Feeling rather silly now... $\endgroup$ – Stefan Dawydiak Oct 1 '13 at 22:01
  • 1
    $\begingroup$ @Imbilio I suggest you answer your own question so this doesn't come up as unanswered. $\endgroup$ – Git Gud Oct 1 '13 at 22:30
  • $\begingroup$ To see a proof based on Daniel Fischer's hint, see this answer to a duplicate question. $\endgroup$ – Martin Sleziak Jun 21 '16 at 20:57
2
$\begingroup$

My hint:

$$L=\lim_{(x,\: y)\to (0,\: 0)}\frac{x^3y^2}{x^4+y^4}=\lim_{(x,\: y)\to (0,\: 0)}\frac{x^2y^2}{x^4+y^4}\: .x$$

Because $$\left|\frac{x^2y^2}{x^4+y^4}\right|\leq\frac{1}{2}\to L=0$$

$\endgroup$
0
$\begingroup$

Let $z=\max (|x|,|y|).$ Then $$|x^3 y^2|\leq z^5.$$ And for $(x,y)\ne (0,0)$ we have $x^4+y^4\geq z^4>0$, implying $$0<1/(x^4+y^4)\leq 1/z^4 .$$ So for $(x,y)\ne (0,0)$ we have $$|(x^3 y^2)/(x^4+y^4|\leq z^5(1/z^4)=z.$$

$\endgroup$

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.