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I want to compute the limit $$ \lim_{(x, y) \to (0, 0)} \sqrt{\left|xy\right| }. $$

In the book I'm reading the author states that one can solve this by using polar coordinates:

Let $x=r\cos\left(\varphi\right), y= r \sin\left(\varphi\right)$ and consider $$ \lim_{r \to 0} \sqrt{\left|r\cos\left(\varphi\right) \cdot r \sin\left(\varphi\right) \right| } = \lim_{r \to 0} |r| \sqrt{|\cos\left(\varphi \right)\cdot \sin\left(\varphi\right) | } = 0. $$

I don't see how this procedure is correct because $\varphi$ is fixed in the limit computation, meaning one goes only along straight lines and doesn't consider all paths. Shouldn't the expression rather be \begin{align*} \lim_{\substack{(r, \varphi)\to (0, \theta)\\\theta \in \mathbb{R}}} \sqrt{\left|r\cos\left(\varphi\right) \cdot r \sin\left(\varphi\right) \right| } =\lim_{(r, \varphi)\to \left\{0\right\} \times \mathbb{R}} |r| \sqrt{|\cos\left(\varphi \right)\cdot \sin\left(\varphi\right) | } = 0 .\end{align*} The result stays the same in this case since the expression involving the squareroot is bounded, but I don't think that the other method is valid in general.

TaChu
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  • I suggest you don't take that book too seriously. It should add another piece of justification: $\sqrt{|xy|}=r\sqrt{|\cos\phi\sin\phi|}\leq r$ (as you mention in your last paragraph). This last inequality gives you a bound independent of $\phi$, and thus we're not making any assumptions about straight lines. But you're right to be cautious; see this answer for more details. – peek-a-boo Sep 29 '21 at 15:14
  • @peek-a-boo One question just arose to me: Suppose that I use the authors method with the constraint that if this method yields a limit value that involves $\varphi$ one can conclude that the limit doesn't exist. Would this method be valid then? – TaChu Sep 29 '21 at 15:22
  • the fact that the limit depends on $\theta$ means the limit's value is affected by which line/path one approaches the origin, so indeed, the limit does not exist in this situation. In any case, I would just suggest avoiding polar coordinates for such problems (the usual cartesian coordinates, and basic inequalities like triangle, AGM, Cauchy-Schwarz are more than enough to deal with such simple functions); they don't really simplify anything and can easily lead to erroneous conclusions. – peek-a-boo Sep 29 '21 at 15:24
  • @peek-a-boo hm I don't really agree. At least for the exercises I have worked on so far the change to polar coordinates simplified things tremendously. Do you have a method that is similarily quick? (I understood your comment as meaning for general limits. I agree that the above example is very simple and hence polar coordinates aren't really needed here) – TaChu Sep 29 '21 at 15:27
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    btw the expression $\lim\limits_{(r,\phi)\to {0}\times \Bbb{R}}$ is not standard, so I'd avoid it (what you probably meant is that the limit is uniform with respect to $\phi$) – peek-a-boo Sep 29 '21 at 15:27
  • @peek-a-boo yes, I changed it in the first but the second somehow didn't change. – TaChu Sep 29 '21 at 15:28
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    Use the trivial estimates $|x|\leq |(x,y)|$ and $y\leq |(x,y)|$ to get $\sqrt{|xy|}\leq \sqrt{|(x,y)|^2}=|(x,y)|$ (i.e it's the same thing as $\sqrt{|xy|}\leq r$, but you don't need the detour into polar coordinates). There are many other such trivial estimates (see the comments of my linked answer for example) – peek-a-boo Sep 29 '21 at 15:29
  • yes, I misunderstood your comment. Thanks a lot for the clarification! – TaChu Sep 29 '21 at 15:29

2 Answers2

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Hint. Please, consider this: We need $(x,y)\to (0,0)$. So we need $(x,y)$ as close to $(0,0)$ we want. That is $(x,y)\in \{(x_1,x_2): ||(x_1,x_2)||<\delta \}=D_{(0,0)}(\delta)$, for appropriately small $\delta >0$. How can we rewrite $D_{(0,0)}(\delta)$ via polar coordinates? We can write $D_{(0,0)}(\delta)=\{(r,φ): r<\delta, φ\in \Bbb R\}$ (it is enough to get $φ\in [0,2π]$). Provided that we have every point of $D_{(0,0)}(\delta)$, when $(x,y)\in D_{(0,0)}(\delta)$, we can obtain any path round $(0,0)$ and inside $D_{(0,0)}(\delta)$ by sketching the line (=path) we want over the points of disc that we want.-

SK_
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The answer to your question is No, and the book answer is perfect: the limit is independent of $φ$, because we need the points $(x,y)$ approaching $(0,0)$ from every path. This can be true only when the polar radius goes to zero and the polar angle is anything. If you force the angle $φ$ to approach zero, you get some linear paths, these which tend to lie on $x$-axis.

SK_
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  • Polar angle being anything does not mean any path. – Matcha Latte Sep 30 '21 at 19:31
  • For any φ and for r close to 0 you get any point round (0,0). That means you can get any point inside a disc of centre (0,0) and small radius (φ is rotating anywhere and radius depends on r), which means that you can draw any line inside this disc -"any line" means "any path". A sketch will help. – SK_ Oct 01 '21 at 07:58
  • Radius of the ball around $(0,0)$ cannot depend on $r$! You should first take a radius that works for every $\varphi$ and every $r$, i. e., the limit should be uniform! What about the paths that are curves? In this case, the polar coordinates might work, but any angle does not mean any path! Polar coordinates are usually used to find a path that is a counterexample so as to prove the limit does not exist. – Matcha Latte Oct 01 '21 at 08:48
  • Of course, if the limit is $0$, it follows that $(x,y)\to(0,0)$ from every path, but you cannot go from the conclusion. You should, at least, mention, $|\cos(\varphi)\sin(\varphi)|\le 1$. Otherwise, the OP will think you are indeed treating $\varphi$ as a constant. You cannot use the consequence of a result to prove a result. – Matcha Latte Oct 01 '21 at 09:46
  • Please, see the added hint above the answer. – SK_ Oct 01 '21 at 10:36