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How can I apply the squeeze theorem to show $$\lim_{(x,y)\to (0,0)} \frac{xy^5}{x^4+y^6}=0$$

amWhy
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Probability student
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    How have you applied the squeeze theorem? – Parcly Taxel Mar 09 '18 at 11:02
  • Welcome to MSE. It will be more likely that you'll get an answer if you show us that you made some effort. – José Carlos Santos Mar 09 '18 at 11:06
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    Take the equivalent expression $$\frac{1}{\frac{x^3}{y^5}+\frac yx}$$ – Piquito Mar 09 '18 at 11:16
  • try it with $$y=tx$$ – Dr. Sonnhard Graubner Mar 09 '18 at 11:19
  • Well trying with $y=tx$ gives the claimed limit, but it does not show that the limit is unique. There are some examples where approaching by any line gives some unique limit, but approaching with another curve (quadratic for instance) gives another one... – Tal-Botvinnik Mar 09 '18 at 11:35
  • @Dr.SonnhardGraubner Why propagate this ultra classical misconception? Because you really think that if $f(x,tx)\to0$ when $x\to0$, for every $t$, then $f(x,y)\to(0,0)$ when $(x,y)\to(0,0)$? Exercise for you: Find a counterexample. – Did Mar 09 '18 at 12:35
  • @Piquito What's next? – Did Mar 09 '18 at 12:39
  • I think the order of increase of $y^5$ being greater than that of $x^3$ the quotient gives infinite so whatever be the quotient $\dfrac yx$ you have $\infty$ in the denominator. – Piquito Mar 10 '18 at 10:58
  • @Piquito Sounds awfully vague, no? – Did Mar 10 '18 at 21:13
  • I understand it could "sounds awfully vague" for many people; not for me. Regards. – Piquito Mar 11 '18 at 14:54
  • @Piquito Care to explain what you meant in this first comment then, for the understanding of people who are not yourself? (Unrelated: Please use @, unless you want your comments to be missed by the user they are addressed at.) – Did Mar 11 '18 at 19:52
  • There are levels of knowledge, that's all. – Piquito Mar 12 '18 at 16:48
  • @Piquito Sorry but you are not making any sense. No big deal. – Did Mar 19 '18 at 19:41

2 Answers2

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If $|x|\geqslant|y|^{4/3}$ and $x\ne0$, $$\left|\frac{xy^5}{x^4+y^6}\right|\leqslant\left|\frac{xy^5}{x^4}\right|=\frac{|y|^5}{|x|^3}\leqslant|x|^{3/4}$$ If $|x|\leqslant|y|^{4/3}$ and $y\ne0$, $$\left|\frac{xy^5}{x^4+y^6}\right|\leqslant\left|\frac{xy^5}{y^6}\right|=\frac{|x|}{|y|}\leqslant|y|^{1/3}$$ Thus, for every $(x,y)\ne(0,0)$, $$\left|\frac{xy^5}{x^4+y^6}\right|\leqslant|x|^{3/4}+|y|^{1/3}$$ The limit of the LHS when $(x,y)\to(0,0)$ follows.

Did
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Hint: for $xy = 0$ the expression is zero, for $xy\ne 0$ bound the denominator using the weighted AGM inequality with weights 1 and 3: $$|x||y|^{9/2}\le\frac{x^4 + 3y^6}4\le x^4 + y^6.$$

Martín-Blas Pérez Pinilla
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  • but this doesnt show that the limit is zero. Observe that the RHS can define any non-negative value for a chosen path $(x,y)\to(0,0)$ – Masacroso Mar 09 '18 at 11:29