Proving Limit of (x^2+y^2)sin(1/(x^2+y^2))→0 as (x,y)→(0,0)

[Nick89]

Homework Statement

Show that
$$\displaystyle \lim_{(x,y) \to (0,0)} (x^2+y^2) \sin \left( \frac{1}{x^2+y^2} \right) = 0$$

This question came up in an exam and I want to see if I got it right... I am doubtful though since I know limits of two variables often don't exist (because they have to be unique for each approach).

The Attempt at a Solution

I reasoned that the argument of the sine will tend to infinity, but the sine itself will still always stay between -1 and 1. Because the 'pre-factor' (x^2 + y^2) goes to 0, this limit is 0.

I am pretty sure that my answer is not 'valid enough', because I'm sure the objective here was to proof it formally using epsilons and delta's... But I don't understand how I would do that here...

Can you tell me what you think? Is this proof formal enough or should I have proven it more carefully? I hope you understand what I mean...

 

[jostpuur]

It is unfortunate truth that sometimes it is unclear that how detailed you need to get in the proof. Sometimes the exercise or an exam question states something like "prove by using the definition" or something else to emphasize what you need to do. If the task had been to determine

$$\lim_{(x,y)\to 0} (x^2+y^2)\sin\Big(\frac{1}{x^2+y^2}\Big),$$

one might think that your argumentation would have been well sufficient, but if the task was to show that

$$\lim_{(x,y)\to 0} (x^2+y^2)\sin\Big(\frac{1}{x^2+y^2}\Big) = 0,$$

then it would seem natural to use the definition of the limit.

Nobody on the physicsforums can promise you what would have been right in the exam, but in case you are interested, we can help you to prove the theorem, that if a function $$f:\mathbb{R}^n\to\mathbb{R}$$ has the limit $$f(x)\to 0$$ when $$x\to 0$$, and if a function $$g:\mathbb{R}^n\to\mathbb{R}$$ is bounded in some environment of the origo, then $$f(x)g(x)\to 0$$ as $$x\to 0$$. First write down the definitions to figure out what you are supposed to prove, and then the key is to use number

$$\underset{x\in U}{\textrm{sup}} |g(x)| < \infty$$

where $$U\subset\mathbb{R}^n$$ is some environment of origo.

 

[gamesguru]

Your argument is perfectly valid and is suitable unless the question otherwise specifies to prove using epsilons and deltas.

 

[HallsofIvy]

The simplest thing to do is to let 1/u= x²+ y² so the problem is
[tex]\lim_{r\rightarrow \infty} \frac{sin(u)}{u}[/itex]
Now you can use your argument that the numerator is bounded while the denominator is goes to infinity.

 

[Nick89]

Well what I meant is probably best to show by example:

Suppose I need to find the following limit:
$$\displaystyle \lim_{(x,y) \to (0,0)} \frac{2x^2y}{x^4+y^2}$$

Let's try a few different approaches:
For example, if you try approaches via the y or x axis, or for that matter by any line y = kx, we get:
$$f(x,kx) = \frac{2kx^3}{x^4+(kx)^2} = \frac{2kx}{x^2+k^2} \to 0$$ as x -> 0.

We might conclude now that the limit was 0. But if we try a different approach, let's say along the curve y = x^2, we get a limit of 1 so it is undefined.

This is what I mean when I say that we really have to prove the limits instead of simply giving a few examples where the limit is the same...

I don't know if this 'few examples' approach is really what I did (I don't think so) but still, it doesn't feel 'correct' in a way... Thanks though!

 

[Defennder]

Werent you just making use of the theorem that if lim x-> a g(x)f(x) = lim x->a g(x) lim x->a f(x)? And if one of them were to converge to 0, the other, loosely speaking must diverge if fg is not to converge to 0, but this is not possible if it were bounded. That's your reasoning isn't it? I don't see anything wrong with that, but then again I'm not a math major so I guess it depends really on how you phrased your reasoning.

 

[jostpuur]

Werent you just making use of the theorem that if lim x-> a g(x)f(x) = lim x->a g(x) lim x->a f(x)?

No.

And if one of them were to converge to 0, the other, loosely speaking must diverge if fg is not to converge to 0, but this is not possible if it were bounded.

Yes. (But that's loosely speaking.)

 

[Nick89]

Well the exact question stated a function f(x,y):
$$f(x,y) = (x^2+y^2) \sin \left( \frac{1}{x^2+y^2}\right) \text{ if } (x,y) \neq (0,0)$$
$$f(0,0) = 0$$

Then the question was to show that this function is continuous in (0,0).
If I got the definition right, a function is continuous at a point (a,b) if
$$\displaystyle \lim_{(x,y) \to (a,b)} f(x,y) = f(a,b)$$

So I have to show that the limit is 0, right?

So the question doesn't explicitly ask me to prove it, but as far as I know, we were taught to prove the limit formally whenever we needed to calculate it.
In the case of one variable only, showing that the limit exists (or is some value) is much easier because there are only two ways to approach the limit (left or right). In the two variables case there are infinitely many ways to approach the limit, and only calculating a few of them of course is not good enough...

 

[HallsofIvy]

It is sufficient to put the problem into polar coordinates and then show that the limit does not depend on the angle $\theta$. In polar coordinates, the distance from (0,0) is measured only by r, not $\theta$. That's essentially what I suggested above.