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Im trying to prove that the function $$\begin{cases}f(x,y)=\dfrac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}, & (x,y)≠0\\ 0, & (x,y)=0\end{cases}$$ is continuous at point (0,0) using the rigorous defintion of a limit.

Attempting to find the upper limit of the function: $$|f(x)-f(x_0)|= \left|\frac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}-0\right|$$ I see the denominator is always positive so this is equal to $\dfrac{|2x^2y^4-3xy^5+x^6|}{(x^2+y^2)^2}$. Using the triangle inequality i know that this is equal or less than $\dfrac{|(2x^2y^4)-(3xy^5)|+|x^6|}{(x^2+y^2)^2}$. From here I would like to continue finding expressions which are equal or greater than this, which allow me to cancel some terms against $((x^2+y^2)^2)$. Im thinking i can write $$x^6 = (x^2)^3 ≤ (x^2+y^2)^3 $$ for instance, but i am unsure of how to "handle" $|(2x^2y^4)-(3xy^5)$|. Could someone give me any pointers?

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  • $\begingroup$ See here for a guide to formatting on this site. Basically, put mathematical expressions between $s. If you want superscripts, use ^ as you have been. To get subscripts, use underscores, e.g. $x_0y^2$ produces $x_0y^2$. To get fractions, you can either use / as you have been doing, or use \frac{}{}. The curly braces are for "grouping". For example $\frac{x^2+y^2}{x+y}$ produces $\frac{x^2+y^2}{x+y}$. I would help, but I'm not sure how to parse 2x2y^4. $\endgroup$ Sep 9, 2022 at 14:08
  • $\begingroup$ I will use the guide next time. Thank you for formatting $\endgroup$ Sep 9, 2022 at 14:20
  • $\begingroup$ Well, it seems like you know LaTeX formatting anyway (judging by your \lefts, \rights and \begin{cases}s), and this is basically the same. You probably won't need it. :) $\endgroup$ Sep 9, 2022 at 14:21

3 Answers 3

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You can just use : $x \leq \sqrt{x^2+y^2} =r$ and $y \leq \sqrt{x^2+y^2} =r $

Thus $$ 0 \leq | f(x,y) | \leq \frac{2r^6+3r^6+r^6}{r^4} = 6r^2 $$

Then just make $r \to 0$ and it shows that $f(x,y) \to 0$ when ${(x,y) \to (0,0)}$

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Using the triangle inequality like this: $$|f(x,y)|\le \frac{|2x^2y^4|+|-3xy^5|+|x^6|}{(x^2+y^2)^2}$$ Notice that $$\frac{|x^6|}{(x^2+y^2)^2}\le \frac{|x|^6}{x^4}=\frac{x^6}{x^4}=x^2$$ $$\frac{|2x^2y^4|}{(x^2+y^2)^2} \le \frac{2x^2y^4}{y^4}=2x^2$$ $$\frac{|-3xy^5|}{(x^2+y^2)^2} \le \frac{3|x||y|y^4}{y^4}=3|x||y|$$ It is $|f(x,y)| \le 3(x^2+|x||y|)$.

So, noticing that $\sqrt{x^2+y^2}<\delta$ implies $|x|<\delta$ and $|y|<\delta$, you shold be able to conclude with an appropriate choice of $\delta$.

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  • $\begingroup$ Ah, i didnt think of that. Thank you! $\endgroup$ Sep 9, 2022 at 14:19
  • $\begingroup$ @Very_Unwise You're welcome. And welcome to MathStackExchange as well! If you like any answer, you can upvote it and you can accept the one you think is the best pressing the ✓ button on its left, if it has solved your problem. $\endgroup$
    – Bernkastel
    Sep 9, 2022 at 14:34
  • $\begingroup$ "noticing that √(x2+y2)<δ implies |x|<δ and |y|<δ" Using this i attempt to use the proof: Let ε >0 and 𝛿 = √(ε/6). ∣∣∣(2x2y4−3xy5+x6)(x2+y2)2∣∣ ≤ 3(x2+|x||y|) < 6*(√ε/6)^2 < ε $\endgroup$ Sep 9, 2022 at 15:23
  • $\begingroup$ @Very_Unwise that's correct. Good job! $\endgroup$
    – Bernkastel
    Sep 9, 2022 at 15:27
  • $\begingroup$ Huray! Again, thanks for the help :) $\endgroup$ Sep 9, 2022 at 15:28
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Use the fact that $|xy| \le \frac{x^2 + y^2}{2}$. You can then get, $$3|xy^5| = 3|xy|y^4 \le \frac{3}{2}(x + y)^3$$ and $$2x^2y^4 = 2|xy|^2y^2 \le \frac{1}{2}(x + y)^3.$$ You can then proceed as you were.

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