I am having a lot of confusion in my multivariable calculus lectures. We usually see that we can try to approach a certain limit with paths and if two paths yield different "limits", then the limit does not exist. But finding "limits" via any number of different paths does not guarantee the limit exist, in previous lectures, it was not clear if polar coordinates were a considered a kind of path, but the professor didn't say it wasn't.

It was said that only the $\epsilon-\delta$ definition actually guarantees the limit exist. In a local calculus book, I found the following theorem (which was also presented in previous calculus lectures):

  • If $\displaystyle \lim_{(x,y)\to {(x_0,y_0)}}f(x,y)=0$ and $|g(x,y)|\leq M$ for $0<|(x,y)-(x_0,y_0)|<r$ with $r>0, M>0$ fixed real numbers, then:

$$\lim_{(x,y)\to {(x_0,y_0)}}f(x,y)g(x,y)=0$$

I tried to use it in the following two cases:

  • $$\lim_{(x,y)\to (0,0)} \frac{xy}{\sqrt{x² + y²}} $$

Here we see that:

$$\lim_{(x,y)\to (0,0)} \frac{xy}{\sqrt{x² + y²}} = x \frac{y}{\sqrt{x² + y²}} $$

And it's easy to see that $\frac{y}{\sqrt{x² + y²}}$ is bounded and we can use the previous thorem, also Wolfram Alpha asserts it exists.

  • $$\lim_{(x,y)\to (0,0)} x \sin \left( \frac{1}{x²+y²} \right)$$

Now, it seems a lot that $\left |\sin \left( \frac{1}{x²+y²} \right)\right|\leq 1$ and that we can apply our theorem, but Wolfram Alpha says the limit does not exist.

Now, as I said before, it was said that approaching via any number of paths does not prove that a limit indeed exists and it was not clear wheather the approach via polar coordinates is considered a "path" (although, to me, it looks like something very diferent than the paths $y=mx$, for example), now there is a new professor who gave us a list of exercises with the following proposition:

  • Given $f: D \subset \Bbb{R}² \to \Bbb{R}$, then

$$\lim_{(x,y)\to0 } f(x,y)=L \iff \lim_{r\to0} f(r\cos \theta,r \sin \theta)=L$$ uniformly in $\theta$. That is, the limit exists iff the radial limit of $f$ does not depend on $\theta$.

But what is "uniformly"? What is "does not depend"? I have tried to use it on the second example, it gives me:

$$\lim_{r\to 0} r \cos \theta \sin\left(\frac{1}{r²}\right) $$

And then I assumed that "does not depends" means that no part of the formula contains $\theta$. But upon inspecting my first example, it gives me:

$$\lim_{r\to 0} r\cos \theta \sin \theta $$

If I assume this meaning for "does not depend", then this limit does not exist, but the other theorem asserts it does and I have also verified on Wolfram Alpha. I have talked with this professor, she said me that the theorem on radial limits is true and that she saw it in a book (I never saw it anywhere), I asked her the title of the book and am still waiting a response, she also said that the polar coordinates are not really a "path". From this, I have these questions:

  1. Is the first theorem actually a theorem?

  2. Is the second theorem actually a theorem? Where can I find a proof of it? I googled a bit something like "polar coordinate limits" and "radial limits" but didn't find anything that seemed relevant. I am asking if it is true because if it is, I still can't prove it.

  3. If the first theorem is true, then where am I making any mistake? I suspect I may be misreading the condition of the distance bounded by $r$.

  4. What is the meaning of "depend" and "uniformly"?

  5. I have noticed the following possibility (may be wrong): $$\lim_{r\to 0} r \cos \theta \sin\left(\frac{1}{r²}\right) =\cos \theta \lim_{r\to 0} r \sin\left(\frac{1}{r²}\right) $$

and I know that

$$\lim_{r\to 0} r \sin\left(\frac{1}{r²}\right)=0$$

and $|\cos x| \leq 1$, so what could possibly go wrong if we take a different $\theta$? It really seems that a change in $\theta$ won't change the limit.

  • $\begingroup$ WolframAlpha does not compute multivariate limits. $\endgroup$ Commented Aug 29, 2018 at 8:32
  • $\begingroup$ @JuliánAguirre No? $\endgroup$
    – Red Banana
    Commented Aug 29, 2018 at 14:02
  • $\begingroup$ That is an iterated limit. $\endgroup$ Commented Aug 29, 2018 at 14:41

2 Answers 2

  1. Yes. It is the squeeze theorem in two variables.

  2. Yes. The idea behind this theorem is that $r:=\sqrt{x^2+y^2} \to 0 \Leftrightarrow (x,y) \to (0,0)$. I have no reference to a proof at the moment.

  3. You need to actually evaluate the limit, and then see whether it depends on $\theta$. For example if we're dealing with $$\lim_{(x,y) \to (0,0)} \frac{x y}{x^2+y^2} $$ then using polar coordinates gives $$ \lim_{r \to 0} \frac{r^2 \cos \theta \sin \theta}{r^2}=\cos \theta \sin \theta$$ which does depend on $\theta$.

  4. "depends" means "a function of", and "uniformly" means "for all possible".

  5. You are correct, the limit in question is equal to $0$, uniformly in $\theta$.

  • $\begingroup$ 5. The trouble is that the limit of $\lim_{(x,y)\to (0,0)} x \sin \left( \frac{1}{x²+y²} \right)$ as $(x,y) \to (0,0)$. And the theorem says that if the radial limit exists, then the limit exists, but it doesnt seems to be the case. $\endgroup$
    – Red Banana
    Commented Aug 29, 2018 at 14:16
  1. It seems like a multivariate version for calculation laws of infinitesimals. You might be able to prove it by $\varepsilon-\delta$ definition.
  2. Still could proved by definition. Note that $r = \sqrt {x^2 + y^2}$.
  3. Seems that the second example converges... Maybe WolframAlpha is not correct? I do not know.
  4. Generally $f(r \cos(\theta), r \sin(\theta))$ is a bivariate function $g(r, \theta)$, then if $\lim_{r \to 0} g(r,\theta)$ exists, then the result should be something like $\varphi(\theta)$. The uniform convergence here is an analogy of that in the discussion of series of functions or sequence of functions, which is the content of single-variable calculus [although exceptions exist].
  5. Changing $\theta$ would not change the limit.

For convenience, I post the definition of uniform convergence here:

For a bivariate function $g(x,y)$ defined on some domain $D$ and $(x_0, y_0) \in D$, we say that $g(x,y) \rightrightarrows g(x_0,y)$ [or in words, $g(x,y)$ converges to $g(x_0,y)$ uniformly] as $x \to x_0$ if for every $\varepsilon > 0$, there is some $\delta > 0$ such that for every $y\colon (x,y) \in D$, $|g(x_0,y) - g(x,y)| < \varepsilon$ whenever $|x - x_0| < \delta$.


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