4
$\begingroup$

I need to prove that

$$\lim_{(x,y)\to (0,0)}\frac{xy}{\sqrt{x^2 +y^2}}=0$$

And I have already found some questions with the exact same question but I guess I proceeded in a somewhat different way.


The limit exists if for each $\epsilon>0$, there is $\delta>0$ such that

$$\left|\cfrac{xy}{\sqrt{x^2+y^2}} -L\right|<\epsilon$$

And $|x-a|<\delta,|x-a|<\delta$ or $|(x,y)-(a,b)|<\delta$ and $x\neq a, y \neq b $.

I made a somewhat experimental approach and would like to talk about it. I learned that we can try to approach the limit with some paths: polar coordinates, lines, etc. So I give a shot: As the limit goes to $(0,0)$, it seemed natural to try: $x \to \frac{1}{m} $ as $m \to \infty $, and $y\to \frac{1}{n}$ as $n\to \infty$, this gives me:

$$\left|\cfrac{1}{\sqrt{m^2+n^2}}\right|<\epsilon$$

$$\left|\frac{1}{m}\right |<\delta \quad \quad \left|\frac{1}{n}\right|<\delta $$

And with this, I guess that we can see that we can approach $0$ arbitrarily. But if we take the other form:

$$\left|\sqrt{\frac{1}{m^2}+\frac{1}{n^2}}\right|<\delta$$

$$\left|\sqrt{\frac{m^2+n^2}{m^2n^2}}\right|<\delta$$

$$\left|\frac{\sqrt{m^2+n^2}}{ mn} \right|<\delta$$

But now it seems that we can't have a $\delta$ for every $\epsilon$ chosen. So how can they be equivalent? I may be missing something. Also, one important question:

Q: I learned the definitions with $\epsilon, \delta$ and the idea of looking for a possible limit by some paths as different ways: The first one seems to be used to prove that there is a limit, but the second one seems to be a cheap method to gather evidence that it couldn't be (if the value is different for two paths, then the limit does not exist), but in the above example, I merged them both. Is this usually viable or do I incur the same problem of not having the same value for different paths implies the non-existence of the limit?


I had also the (incomplete?) idea of trying to approach the limit with a circle $a^2+b^2=c^2$ and take the limit of $c\to 0$ but am a little confused if it could be done.

$$\left|\frac{\sqrt{c^2-a^2} \sqrt{c^2-b^2}}{\sqrt{-a^2-b^2+2 c^2}}-0\right|<\epsilon$$

$$\left|\frac{\sqrt{c^2-a^2} \sqrt{c^2-b^2}}{\sqrt{-a^2-b^2+ c^2+c^2}}\right|<\epsilon\tag{$c^2-a^2-c^2=0$}$$

$$\left|\frac{\sqrt{c^2-a^2} \sqrt{c^2-b^2}}{c}\right|<\epsilon$$

As $b^2=c^2-a^2$ and $a^2=c^2-b^2$, then:

$$\left|\frac{ba}{c}\right|<\epsilon$$

As $c\to 0$, I guess we can make the substitution $c:= \frac{1}{n}$ with $n\to \infty$, then:

$$\left|abn\right|<\epsilon$$

And here, as $n\to \infty $, $c\to 0$ and due to $a^2+b^2=c^2$, then $a,b\to 0$. As for the $\delta$, then:

$$\left|\left( \sqrt{c^2-a^2},\sqrt{c^2-b^2} \right) -(0,0)\right|<\delta$$

$$\left| \left( \sqrt{c^2-a^2+c^2-b^2 }\right)\right|<\delta\tag{$c^2-a^2-c^2=0$}$$

$$|c|<\delta$$

Are these moves acceptable? If not, do you know at least one counterexample? I'd like to see why it doesn't work - if it doesn't work.

$\endgroup$

4 Answers 4

2
$\begingroup$

I will read your solution carefully, but for sake of something I think is very quick, change to polar coordinates.

$$\lim_{r \to 0} \frac{r^2 \sin \theta \cos \theta}{r} = 0$$

Edit: Yes, I am being a bit sloppy here. As written, it looks as though I am assuming $\theta$ is a fixed constant, but it can be a function of $r$.

$$\lim_{r \to 0} r \sin \left(\theta(r)\right) \cos \left(\theta(r)\right) \leq \lim_{r \to 0} M r = 0 $$

for some positive $M$.

$\endgroup$
2
  • $\begingroup$ Yes, I've seen that solution. But I decided to wrestle with the scary $\epsilon$'s and $\delta$'s. So I am trying some ideas. $\endgroup$
    – Red Banana
    Mar 22, 2017 at 2:41
  • $\begingroup$ This doesn't answer the question. It's just another solution. $\endgroup$ Mar 22, 2017 at 3:17
1
$\begingroup$

I will quote here the definition of the limit of a 2-variable function: $$\forall \epsilon>0(\exists\delta(\|(x-a,y-b)\|\leq\delta\implies\|f(x,y)-L\|)\leq\epsilon)$$ that is, for all positive real value $\epsilon$, there exists a circle (of radius $\delta$) inside which the function's value is not further away from $L$ than $\epsilon$. Therefore no matter how (by which curve) you approach the point $(a,b)$, once you enter the circle, you are less-than-$\epsilon$-close to $L$. That's why to show that the limit doesn't exist, you only have to find one path for which the condition is not satisfied. To show that the limit does exist using the path approach, you would have to prove that for all paths arriving at $(a,b)$, the condition is satisfied. We cannot do this by considering all the paths, since only the classification of these paths is complicated enough.

I would like to add to Faraad Armwood's answer the fact that $\sin\theta\cos\theta$ is a bounded function.

$\endgroup$
1
$\begingroup$

You solution is essentially equivalent to Faraad's where $c$ plays the roles of $r$.

Since you are playing with $\epsilon$ and $\delta$, you need to show that for any given $\epsilon$, there is a $\delta$ so that $|f(x,y)-L|<\epsilon$ whenever $c<\delta$. In other words, you need to show how $\epsilon$ and $\delta$ relate, which is lacked in your solution. This is also mentioned in Mr.T's answer.

To sum up, your solution works when you choose (add?) $\delta = \epsilon$. That is, $$\left|\frac{ab}{\sqrt{a^2+b^2}}\right|=\left|\frac{ab}{c}\right| \le |c| \le \epsilon,$$ whenever $|c|\le \epsilon$.

$\endgroup$
1
$\begingroup$

By the GM-AM inequality,

$$|xy|\le (x^2+y^2)/2 \le x^2+y^2.$$

So in absolute value the expression of interest is bounded above by $\sqrt {x^2+y^2}.$ Since $\sqrt {x^2+y^2} \to 0,$ the desired limit is $0.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .