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For example, when we compute $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y^2}$, we can use the change to polar coordinates method below:

Let $\begin{cases}x=r\cos\theta\\y=r\sin\theta \end{cases}$, then $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y^2}=\lim_{r\to0}\frac{r^3\cos^3\theta\sin\theta}{r^2}(=\lim_{r\to0}r\cos^2\theta\sin\theta=0)$.

However, why is this method valid? What is the actual reason behind this substitution? Is this related to the limit law of composition functions of continuous function? But what are those two functions that applying behind here?

I think it makes use of this change-of-coordinate continuous and one-to-one function $\Psi$:

$$\begin{alignat*}{3}\Psi:\ &\mathbb{R}^+\times[0,2\pi)&&\to\mathbb{R}^2\\ &(r,\theta)&&\mapsto(r\cos\theta,r\sin\theta)\end{alignat*}$$

But I can't figure out how to "chain" this function with others and get the eager result.

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  • $\begingroup$ the important thing is that you approach the same point, on the same function, we are using equality and r is a continuous limit to a specific and unique point which the path is also uniquely by equality of x=rcos(theta) and y=rsin(theta) bound to the original "path" equation defined the limits then cannot help but work. $\endgroup$ – shai horowitz Oct 30 '16 at 6:25
  • $\begingroup$ @shaihorowitz The intuition arguement is difficult for me. Can you write down in symbolic manner expressions? $\endgroup$ – Eric Oct 30 '16 at 6:33
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Note: this answer stems from the (now removed) comments left under my answer to your other question about polar coordinates. As requested, I have edited to add the proofs.

Before proceeding, let us define (possibly non-standard notation) $$\lim_{(r,\phi)\to\{0\}\times\mathbb R}g(r,\phi)=L\tag{$\diamond$}$$ to mean that for each $\epsilon>0$, there is an open neighborhood $U$ of $\{0\}\times\mathbb R$ in $[0,\infty)\times\mathbb R$ such that $(r,\phi)\in U$ and $r\neq0$ implies $|g(r,\phi)-L|<\epsilon.$

In the case $g(r,\phi)=f(r\cos\phi,r\sin\phi)$, which is $2\pi$-periodic in the second argument, this can be simplified, by requiring instead that for each $\epsilon>0$, there is a $\delta>0$ such that $0<r<\delta$ implies $|g(r,\phi)-L|<\epsilon$. (Intuitively, in this case we do not have to worry about $\delta$ getting smaller and smaller as $\phi\to\pm\infty$.) This can be done e.g. by using the tube lemma from elementary topology.

We may now state:

Proposition. The following are equivalent for a function $f:\mathbb R^2\setminus\{(0,0)\}\to\mathbb R$:

  1. We have $$\lim_{(x,y)\to(0,0)}f(x,y)=L.$$
  2. The function $\tilde f:\mathbb R^2\to\mathbb R$, defined by $$\tilde f(x,y)=\begin{cases}f(x,y);&(x,y)\neq(0,0),\\L;&(x,y)=(0,0),\end{cases}$$ is continuous at $(0,0)$ and $f(0,0)=L$.
  3. For all $\phi_0\in\mathbb R$ we have $$\lim_{(r,\phi)\to(0,\phi_0)}f(r\cos\phi,r\sin\phi)=L.\tag{$\star$}$$
  4. We have $$\lim_{(r,\phi)\to\{0\}\times\mathbb R}f(r\cos\phi,r\sin\phi)=L.$$
  5. The function $g:[0,\infty)\times\mathbb R\to\mathbb R$, defined by $$g(r,\phi)=\begin{cases}f(r\cos\phi,r\sin\phi);&r>0,\\L;&r=0,\end{cases}$$ is continuous at $(0,\phi_0)$ for each $\phi_0\in\mathbb R$.

Proof. The equivalence of 1 and 2 is standard and follows directly from the definitions.

To see that 2 implies 5, let $p:[0,\infty)\times\mathbb R\to\mathbb R^2$ be defined by $p(r,\phi)=(r\cos\phi,r\sin\phi).$ This is obviously continuous, so if $\tilde f$ is continous, $g=\tilde f\circ p$ must be continuous as well.

Next, we show that 5 implies 3. To see this, let $\epsilon>0$ and $\phi_0\in\mathbb R$. By $5$, there is a $\delta>0$ such that $\|(r,\phi)-(0,\phi_0)\|<\delta$ implies $|g(r,\phi)-g(0,\phi_0)|<\epsilon$. In particular, if $r>0$, this means that $\|(r,\phi)-(0,\phi_0)\|<\delta$ implies $|f(r\cos\phi,r\sin\phi)-L|<\epsilon$, which is precisely 3.

Next, we show that 3 implies 4. Let $\epsilon>0$. For each $\phi_0\in\mathbb R$ we have a $\delta(\phi_0)>0$ such that $\|(r,\phi)-(0,\phi_0)\|<\delta(\phi_0)$ implies $|f(r\cos\phi,r\sin\phi)-L|<\epsilon$. Let $$U(\phi_0)=\{(r,\phi)\in[0,\infty)\times\mathbb R\mid \|(r,\phi)-(0,\phi_0)\|<\delta(\phi_0)\}.$$ Then $U=\bigcup_{\phi_0\in\mathbb R}U(\phi_0)$ is an open neighborhood of $\{0\}\times\mathbb R$ in $[0,\infty)\times\mathbb R$ and for each $r\neq 0$ the condition $(r,\phi)\in U$ implies $\|f(r\cos\phi,r\sin\phi)- L\|<\epsilon$, because every such $(r,\phi)$ is an element of some $U(\phi_0)$ and therefore satisfies $\|(r,\phi)-(0,\phi_0)\|<\delta(\phi_0)$.

Finally, we show that 4 implies 1, thus concluding the circle of implications and establishing the desired equivalence. To do this, let $\epsilon>0$. By 4 and the fact that $f(r\cos\phi,r\sin\phi)$ is a $2\pi$-periodic function of $\phi$, there is a $\delta>0$ such that $0<r<\delta$ implies $|f(r\cos\phi,r\sin\phi)-L|<\epsilon$. Therefore, if $0<\|(x,y)\|<\delta$, we have $|f(x,y)-L|<\epsilon$, because $(x,y)=(r\cos\phi,r\sin\phi)$ for some $r\in(0,\delta)$ and $\phi\in\mathbb R$. This concludes the proof. $\qquad\square$

Remark. Note that the limit $(\star)$ is the limit of a two-variable function. Requiring that we have the single variable limit $$\lim_{r\to0}f(r\cos\phi_0,r\sin\phi_0)=L\tag{$\ast$}$$ for each $\phi_0\in\mathbb R$ is not (!) equivalent. In fact, this latter condition is not sufficient for the limit to exist as shown by the standard counterexample $f(x,y)=\frac{x^2y}{x^4+y^2}.$) The point is that the limit $(\star)$ takes into account points $(r,\phi)$ with different values of $\phi$ whereas in $(\ast)$ the value $\phi=\phi_0$ is fixed.

Added: here are the details regarding the part with the tube lemma, but I will do it without actually using the tube lemma, which would allow us to skip a few steps. (In fact, what follows basically includes a proof of the tube lemma in this special case.)

The definition of $(\diamond)$ mentions a neighborhood $U$ of $\{0\}\times\mathbb R$ in $[0,\infty)\times\mathbb R$. We wish to show that, if $g(r,\phi)$ is $2\pi$-periodic in $\phi$, we can always take a neighborhood of the form $V=[0,\delta)\times\mathbb R$ for some $\delta>0$, instead of this possibly more complicated neighborhood $U$.

Assume we have a neighborhood $U$ of $\{0\}\times\mathbb R$, as in the definition of $(\diamond)$, not necessarily of the form $[0,\delta)\times\mathbb R$. Using $U$, we are going to construct a neighborhood $V=[0,\delta)\times\mathbb R$ satisfying the same properties. Note that since $U$ is a neighborhood of every point $(0,\phi_0)\in\{0\}\times[0,2\pi]$, it contains a ball $$B((0,\phi_0),\eta(\phi_0))=\{(r,\phi)\in[0,\infty)\times\mathbb R\mid \|(r,\phi)-(0,\phi_0)\|<\eta(\phi_0)\}$$ of radius $\eta(\phi_0)>0$ around every such point. This ball contains a set of the form $$V(\phi_0)=[0,\delta(\phi_0))\times(\phi_0-\delta(\phi_0),\phi_0+\delta(\phi_0))$$ which still contains $(0,\phi_0)$. The sets $V(\phi_0)$ form a cover of the interval $\{0\}\times[0,2\pi]$, which is compact, so there is a finite subcover $V(\phi_1),V(\phi_2),\ldots, V(\phi_n)$, where $\phi_1,\phi_2,\ldots,\phi_n\in[0,2\pi]$. Now take $\delta=\min\{\delta(\phi_1),\delta(\phi_2),\ldots,\delta(\phi_n)\}$. Observe that $$[0,\delta)\times[0,2\pi]\subseteq V(\phi_1)\cup V(\phi_2)\cup\ldots\cup V(\phi_n)\subseteq U,$$ so $(r,\phi)\in[0,\delta)\times [0,2\pi]$ implies $|g(r,\phi)-L|<\epsilon$. But since $g$ is $2\pi$-periodic in $\phi$, this means that $(r,\phi)\in[0,\delta)\times\mathbb R$ also implies $|g(r,\phi)-L|<\epsilon$. Therefore $V=[0,\delta)\times\mathbb R$ has the desired properties.

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  • $\begingroup$ OK, now I have time back to this question again. I don't know how to clearly show that $\lim_{(x,y)\to(0,0)}f(x,y)=L\Leftrightarrow\lim_{(r,\phi)\to(0,\phi_0)}f(r\cos\phi,r\sin\phi)=L$ by some functions chaining together, and at the final step, how does $g$ come into play?(how to make use of $g$). $\endgroup$ – Eric Dec 3 '16 at 12:43
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    $\begingroup$ I think $\forall \phi_0,~\lim_{(r,\phi)\to(0,\phi_0)}f(r\cos\phi,r\sin\phi)=L$ is intuitively alike that every line through the point would have its functional values goes to $L$. But the latter does not implies $\lim_{(x,y)\to(0,0)}f(x,y)=L$. $\endgroup$ – Eric Dec 3 '16 at 13:14
  • $\begingroup$ Thanks!! I think I understand everything you wrote, except for one tiny thing, that is how to use tube lemma to show that the $\delta$ won't be over small in some tricky situations? I haven't studied topology yet, so I don't know the product-related terminology, but I'm familiar with analysis, so the compactness, openness and so on concepts is OK to me. I think after some struggle I can figure out the details, but I also think there is high probability that I will still mistake something. So maybe you can help me on the details how to apply the tube lemma here. :) $\endgroup$ – Eric Dec 6 '16 at 13:26
  • $\begingroup$ @Eric: Added the proof. $\endgroup$ – Dejan Govc Dec 6 '16 at 14:13
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This might help you: Try to figure out what exactly is the difference between the statements: $$\lim_{r\to0}f(r\cos\theta_0,r\sin\theta_0)=L,\,\forall \theta_0\\ \text{and}\\~\\ \lim_{(r,\theta)\to(0,\theta_0)}f(r\cos\theta,r\sin\theta)=L,\,\forall\theta_0.$$ If you know the difference between the two, then you also know why the first one is pretty much useless, and why the second one is the equivalent to 3. in Andrew D. Whang's answer (and therefore usefull).

As a hint: A drawing is always useful. Your domain looks like this:


The domain

The domain $\big\{(r,\theta)\mid (r,\theta)\in[0,\infty)\times[0,2\pi)\big\}$. The $\color{blue}{\text{blue}}$ line is the line $r=0$.


We want to exeamine the behaviour of $f(r\cos\theta,r\sin\theta)$ as the points $(r,\theta)$ get close to the blue line $r=0$, along any path.

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The right view is not to consider a (or all) "path(s)" leading to $(0,0)$, but the values of the function "near" $(0,0)$ : saying that $\lim_{(x,y)\to(0,0)} f(x,y)=0$ is equivalent to saying $$(\forall\epsilon>0)(\exists\alpha>0) ((||(x,y)-(0,0)||<\alpha)\Longrightarrow (\left|f(x,y)-f(0,0)\right|<\epsilon))$$ for a certain norm $||.||$ of $\mathbb R^2$.

Using polar coordinates is just using the norm $$||(x,y)||_2=\sqrt{x^2+y^2}$$ And the right way to write it is : $$\left|f(x,y)-0\right| = \left|r\cos^2\theta\sin\theta\right|\le r$$ which permits you to use $\alpha=\epsilon$ in the definition (I mean, don't use $\lim_{(x,y)\to(0,0)}$ before you "know" the limit exists).

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  • $\begingroup$ So the reason that $\displaystyle\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y^2}=\lim_{r\to0}\frac{r^3\cos^3\theta\sin\theta}{r^2}$ is not so easy as it looks. It need to back to see the $\epsilon-\delta$ definition to see why it is so? $\endgroup$ – Eric Oct 30 '16 at 7:02
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    $\begingroup$ No, the limits are equal... if they exist !! The majoration makes it evident that they do. $\endgroup$ – Nicolas FRANCOIS Oct 30 '16 at 7:08
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$\newcommand{\Reals}{\mathbf{R}}\newcommand{\eps}{\varepsilon}$Let $f$ be a function defined in some deleted neighborhood $U$ of the origin in $\Reals^{2}$, and let $L$ be a real number. The following are equivalent:

  1. $\displaystyle \lim_{(x, y) \to (0, 0)} f(x, y) = L$.

  2. For every $\eps > 0$, there exists a $\delta > 0$ such that if $(x, y) \in U$ and $0 < \sqrt{x^{2} + y^{2}} < \delta$, then $|f(x, y) - L| < \eps$.

  3. For every $\eps > 0$, there exists a $\delta > 0$ such that if $0 < r < \delta$ and $(r\cos\theta, r\sin\theta) \in U$ for all real $\theta$, then $|f(r\cos\theta, r\sin\theta) - L| < \eps$.

  4. $\displaystyle \lim_{r \to 0} f(r\cos\theta, r\sin\theta) = L$.

Statements 2. and 3. are equivalent for the natural geometric reason that if $(x, y) = (r\cos\theta, r\sin\theta)$, then $\sqrt{x^{2} + y^{2}} = \sqrt{r^{2}} = |r|$, and without loss of generality (e.g., replacing $\theta$ by $\theta + \pi$ if necessary) we may assume $0 < r$.

Edit: Condition 4 is (in my experience) defined by Condition 3. It would be more honest to write something like "$\displaystyle\lim_{r \to 0} f(r\cos\theta, r\sin\theta) = L$ uniformly in $\theta$"; otherwise, this use of limit notation is technically anomalous, since the limit isn't taken at one point, but in a tube neighborhood of the $\theta$-axis. (Of course, the $\theta$-axis maps to one point, the origin, under polar coordinates.)

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  • $\begingroup$ I see!!! So does this usage of polar coordinates essentially HAVE NOTHING TO DO with the limit law of composite functions? For example, this is absolutely not similar to this in any aspect. $\endgroup$ – Eric Oct 30 '16 at 14:35
  • $\begingroup$ The polar coordinates map $(r, \theta) \mapsto (r\cos\theta, r\sin\theta)$ is continuous, and is an open map (which plays the role of "the inverse map is continuous"). Consequently, the fact that "limits at the origin can be studied by converting to polar coordinates" can be viewed as related to composition of continuous maps. It may be simpler, however, to think in terms of the $\eps$-$\delta$ definition. $\endgroup$ – Andrew D. Hwang Oct 30 '16 at 15:07
  • $\begingroup$ I've come back to think this question now. After hours of thinking, I still can't figure out why your 1. is equivalent to 4. either by means of composition of continuous maps that your comment mentioned or by another way. I think the occurence of $\displaystyle \lim_{r \to 0} f(r\cos\theta, r\sin\theta) = L$ is weird for me, what is the role of $\theta$ here? Being constant? $\endgroup$ – Eric Dec 5 '16 at 14:48
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    $\begingroup$ I downvoted because $\lim_{r\to0} f(r\cos \theta , r\sin \theta)=L$ is ill-defined. In fact, I believe it is notation like this that is causing the OP's confusion. $\endgroup$ – gebruiker Dec 5 '16 at 16:25
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    $\begingroup$ @AndrewD.Hwang If you add what you said in the comments to your answer, I can turn my downvote into an upvote. $\endgroup$ – gebruiker Dec 7 '16 at 9:22
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A layman explanation is that the limit exists if and only if you approach the same value no matter which path you take to approach the origin.

A particle on any path which approaches the origin necessarily needs to eventually be constrained in a circle of arbitrarily small radius $r$ centred at the origin. So essentially, with polar coordinates you're "considering" all such possible paths, since you're taking the limit as $r\rightarrow 0$.

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  • $\begingroup$ Can you write down in symbolic manner expressions? (i.e. in detail and clear style) $\endgroup$ – Eric Oct 30 '16 at 7:03

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