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I have to find the values of $n$ where this limit exists or doesn't exist. $$\lim_{(x,y)->(0,0)}\frac{|x||y|^n}{x^2 + y^2}$$

I have tried to use different paths like $y=mx$ or $y=mx^2$ and such but I don' t know what other strategies to try?

suomynonA
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Jacob Boertjes
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    I really dislike using polar coordinates to prove that limits exist. To be honest, I don't understand this method and I think it's wrong. Note that, in case the limits considered below exist, it holds that $$\lim \limits_{(x,y)\to (0,0)}\left(\left|\dfrac{|x||y|^n}{x^2+y^2}\right|\right)\leq \lim \limits_{(x,y)\to (0,0)}\left(\dfrac{|x||y|^n}{|y|^2}\right)=\lim \limits_{(x,y)\to (0,0)}\left(|x||y|^{n-2}\right)_.$$ – Git Gud Jan 19 '17 at 00:00
  • @GitGud Just because you don't like it doesn't mean it's wrong. And post your answer, it is just as justified as any polar method :-) – Simply Beautiful Art Jan 19 '17 at 00:00
  • @SimpleArt But I do think it is wrong. I've never seen a convincing enough (for me) proof that it works. See, for instances, this. – Git Gud Jan 19 '17 at 00:02
  • @GitGud But that one was clearly wrong because there existed $\theta(r)$ such that the denominator approached $0$ as $r\to0$. – Simply Beautiful Art Jan 19 '17 at 00:04
  • @SimpleArt I don't understand the relevance of that. Arguably, the result to use looks something like this: If $\lim \limits_{r\to 0}\left(f(r\cos(\theta),r\sin(\theta))\right)=0$, for all $\theta$ such that the expression makes sense, then $\lim \limits_{(x,y)\to (0,0)}\left(f(x,y)\right)=0$. In the linked example, this statement fails. So I'm guessing that the result isn't exactly this. What is it, then? – Git Gud Jan 19 '17 at 00:17
  • @GitGud One must show that the proposed limit is the case for all $\theta$, which was not done in the link you proposed as a counter-example. For the below, for example, it is the case (due to squeeze theorem), but for your link, if we let$$\theta(r)=\arcsin\left(\frac{-1\pm\sqrt{1+4r^2}}{2r}\right)$$(do excuse me if I made an error, but I think this is right) then the limit is not $0$. So there does exist $\theta$ such that the limit DNE. – Simply Beautiful Art Jan 19 '17 at 00:23
  • @SimpleArt There's no $\theta(r)$. The variable $r$ is a bound variable. One uncommon (but more than enough notation) for limits is $\lim \limits_{0}(g)$. No mention to variables, no mention to $r$, $\theta$ does not depend on $r$. It is true that for all $\theta$ (such that the expression makes sense), $\lim \limits_{r \to 0} \left(\frac{r (\cos^2\theta\sin\theta)}{r^2\cos^4\theta + \sin^2\theta}\right)=0$. – Git Gud Jan 19 '17 at 00:29
  • @GitGud No, it is true for all $\theta$ that are constants that $\lim_{r\to0}=0$. You miss the point that $\theta$ can be a function of $r$, which produces curved paths. – Simply Beautiful Art Jan 19 '17 at 00:31
  • @SimpleArt OK. If you're genuinely interested in helping me understand this, formalize the statement as much as you can. Something like "Given a function $f$ defined on $\mathbb R^2$ and letting $g\colon\mathbb R^2\to \mathbb R^2, (r,\theta)\mapsto (r\cos(\theta), r\sin(\theta))$, consider the function $f\circ g$. If _______ then _______" and complete the blank spaces. To simplify consider the limit at the origin. – Git Gud Jan 19 '17 at 00:36
  • When I look at$$\lim_{(x,y)\to(0,0)}f(x,y)$$I would usually interpret it as "what is the limit of this such that for all $y=g(x)$ or $x=h(y)$ that have $(x,y)\to(0,0)$", we will then have$$\lim_{x\to0}f(x,g(x))=\lim_{y\to0}f(y,h(y))=c$$else, if it fails, then the limit doesn't exist. I'm no expert in this, now am I able to create formal statements (ok, I need to take real analysis or something), but I think my intuition is very strong, and as long as there are relations between $x$ and $y$, there are relations between $r$ and $\theta$. – Simply Beautiful Art Jan 19 '17 at 00:39
  • @SimpleArt That's the problem I've been facing all along. No one is able to formalize this for me. I reject your formulation because you're using the unformalizable concept of "function of a variable". Functions are functions, they are ordered pairs by definition. Notations such as $y=g(x)$ and $x=h(y)$ are simply notational intuitive devices people use to think about things, but which can't be properly formalized (in $\sf ZFC$). Get back to me when you're mathematical apparatus is strong enough to give me what I need $\ddot \smile$ – Git Gud Jan 19 '17 at 00:45
  • @GitGud then I'll try to see you then :-). And thanks for the very intellectual conversation. – Simply Beautiful Art Jan 19 '17 at 00:51

3 Answers3

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Turn it into polar form:

$$\lim_{r\to0}\frac{|\cos(\theta)||\sin^n(\theta)|}{r^{1-n}}=\lim_{r\to0}r^{n-1}|\cos(\theta)||\sin^n(\theta)|$$

So the limit exists for $n>1$.

Simply Beautiful Art
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You need "enough degree" in the numerator to be able to cancel the denominator.

When $n=1+c$, with $c>0$, we have (using that $|x|\leq\sqrt{x^2+y^2}$ and similar for $y$) $$ \frac{|x|\,|y|^n}{x^2+y^2}=|y|^c\,\frac{|x|\,|y|}{x^2+y^2} \leq|y|^c\,\frac{x^2+y^2}{x^2+y^2}=|y|^c\to0. $$ When $n\leq1$, consider the path $x=my^n$, $x,y>0$. Then $$ \frac{xy^n}{x^2+y^2}=\frac{my^{2n}}{m^2y^{2n}+y^2} =\frac{m}{m^2+y^{2-2n}}\to\frac1m. $$ So, for $n\leq1$ the limit along different path is different, and so the limit doesn't exist.

In summary, the limit exists for $n>1$ and does not exist for $n\leq1$.

Martin Argerami
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$$\lim_{(x,y)\to(0,0)}\frac{|x||y|^n}{x^2 + y^2}$$ the limit is readily solved in polar form; using $\begin{cases}x=\rho\sin{\theta}\\y=\rho\cos{\theta}\end{cases}$, one obtains: $$\lim_{\rho\to0^+}\frac{\rho^{n+1}|\sin \theta||\cos \theta|^n}{\rho^2}=\lim_{\rho\to0^+}\rho^{n-1}|\sin \theta||\cos \theta|^n=\begin{cases}\rm{undefined}&n\le1\\0&n>1\end{cases} $$ to have a continuos function the limit has to be defined and independent of $\theta$ hence $n>1$ must be satisfied. if $n=1$ the function admits no limit.

Frank
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