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How do you find the following limit:

$$\lim_{x,y\to 0} \frac{|x||y|^4}{|x|^4+|y|^5}$$

I've tried to apply the squeeze theorem, but unlike a lot of other questions I encounter, the degree of $x$ in the numerator is too small for me to write the function as a constant times a function that goes to zero as $x,y \rightarrow 0$.

user
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Snowball
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2 Answers2

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We can rewrite the expression like so

$$\frac{|x|y^4}{x^4+|y|^5} = \frac{|x|y^4}{\sqrt{x^4+|y|^5}}\cdot\frac{1}{\sqrt{x^4+|y|^5}}$$

By AM-GM inequality (or if you don't want to use AM-GM, simply complete the square) we have that

$$x^4+|y|^5 \geq 2x^2|y|^{\frac{5}{2}}$$

which means

$$\frac{|x|y^4}{\sqrt{x^4+|y|^5}}\cdot\frac{1}{\sqrt{x^4+|y|^5}} \leq \frac{|x|y^4}{\sqrt{2x^2|y|^{\frac{5}{2}}}}\cdot\frac{1}{\sqrt{x^4+|y|^5}} = \frac{|y|^{\frac{11}{4}}}{\sqrt{2}\sqrt{x^4+|y|^5}}$$

and lastly we have that

$$\frac{|y|^{\frac{11}{4}}}{\sqrt{2}\sqrt{x^4+|y|^5}} \leq \frac{|y|^{\frac{11}{4}}}{\sqrt{2}|y|^{\frac{5}{2}}} = \left|\frac{y}{4}\right|^{\frac{1}{4}}$$

thus we have that the limit

$$\lim_{(x,y)\to(0,0)}\frac{|x|y^4}{x^4+|y|^5} \leq \lim_{(x,y)\to(0,0)} \left|\frac{y}{4}\right|^{\frac{1}{4}} = 0$$

by squeeze theorem. As a rule of thumb, if the path substitution that makes the denominator homogeneous (i.e. the sum of terms of the same degree) has the limit exist and agree with the other paths you've tried, then it probably means the limit exists, and more importantly, a dominating function exists.

Ninad Munshi
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    This is a good example for the convenience to use polar coordinates avoiding to find the proper inequality to get the result. Of course, I agree this one is very clever and elegant. – user Aug 22 '20 at 22:21
  • Correct me if I'm mistaken but this answer seems generalizable to any positive exponents? For example:$\lim_{x,y\to 0} \frac{|x||y|^\pi}{|x|^4+|y|^5}$ – Snowball Aug 22 '20 at 22:25
  • @Snowball not any positive exponents, no. If the homogenous limit (in this case the path $y=x^{\frac{4}{5}}$) equaled a constant, or worse blew up, then this limit wouldn't exist. In this case if you only changed the numerator power of $y$, the limit would only exist if the exponent were greater than or equal to $\frac{15}{4}$ so no, $\pi$ doesn't work. – Ninad Munshi Aug 22 '20 at 22:26
  • Sorry, yes, what I mean is that the method is generalizable to positive exponents. The limit exists if and only if $\frac{a_1}{b_1}+\frac{a_2}{b_2} > 1$ where $a_1, a_2$ are the exponents of x,y in the numerator and $b_1$ and $b_2$ are the exponents of x,y in the denominator. – Snowball Aug 22 '20 at 22:35
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We have that by $|y|=|v|^{\frac45}\to 0$

$$\frac{|x||y|^4}{|x|^4+|y|^5} =\frac{|x||v|^{\frac{16}5}}{|x|^4+|v|^4} \to 0$$

as we can conclude by polar coordinates, indeed

$$\frac{|x||v|^{\frac{16}5}}{|x|^4+|v|^4}=r^{\frac15}\frac{|\cos \theta||\sin\theta|^{\frac{16}5}}{|\cos \theta|^4+|\sin \theta|^4} \to 0$$

since $\forall \theta$

$$|\cos \theta|^4+|\sin \theta|^4 \ge c >0$$

user
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    To play devil's advocate, there are functions that appear to have a limit at the origin in polar coordinates but actually do not (such as $f(x,y)=\frac{x^2y}{x^4+y^2}, (x,y)\neq(0,0)$ and $f(0,0)=0$). Why is your use of them justified here? – Integrand Aug 22 '20 at 21:48
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    @Integrand Polar coordinates are a standard method to study multivariable limits. They are very useful indeed or to show limit exists, as in this case, avoiding the needing to find some good inequality which works or to show the limit doesn't exist avoiding the needing to find the right paths. I use that tecnique often but of course we can also find other methods. – user Aug 22 '20 at 21:52
  • @Integrand For your example by $x^2=u$ we obtain $$\frac{x^2y}{x^4+y^2}=\frac{uy}{u^2+y^2}=\cos \theta \sin \theta$$ which indeed has not limit at $(0,0)$. Of course we need to apply methods properly but this is true in general and not only for polar coordinates. – user Aug 22 '20 at 21:55
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    @Integrand Why do you say that the function appears to have a limit at $(0,0)$ if one uses polar coordinates? It seems just the opposite to me. Won't it behave like $\frac1r$? – saulspatz Aug 22 '20 at 21:57
  • @saulspatz look at the cases $y=0$ and $y\ne 0$ separately – Integrand Aug 22 '20 at 22:03
  • @Integrand the only caveat is that the limit be "uniform" in $\theta$ (and this uniformity cannot be achieved by taking cases, which is what happens in the counterexample you provide assuming we use the variable substitution $y=v^2$ like what user has done here to make the denominator homogeneous) – Ninad Munshi Aug 22 '20 at 22:07
  • @NinadMunshi Of course if we apply the method wrongly very often the answer obtained will be wrong! That's fine. I don't understand what is the counterexample presented here. – user Aug 22 '20 at 22:12
  • @user that was an explanation for integrand, I know you understand it – Ninad Munshi Aug 22 '20 at 22:17
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    To add to your answer, you can use that $c = \frac{1}{2}$ – Ninad Munshi Aug 22 '20 at 22:18
  • @NinadMunshi I mean that homogenity is not a limitation for using polar coordinates but a clever way to obtain the desired result. Thanks for your intervetion to clarify better the issue. – user Aug 22 '20 at 22:19
  • Also I think I see the confusion. I meant to say "we don't use" instead of "use" the variable substitution – Ninad Munshi Aug 22 '20 at 22:21