# How to find this limit?

1. Mar 4, 2006

### pivoxa15

Find the limit as (x,y)->(0,0) for (yx^2)/(x^2+y^2)

If do this literally, I get 0/0 hence no limit but a limit does exist for this function. How do I get it? What is the general way to find the limits of multi variable functions?

2. Mar 4, 2006

### VietDao29

Ehh???
Who tells you that 0 / 0 means the limit does not exist there? It's one of the Indeterminate forms.
Generally, to find a limit of a 2 variable function, one can change it to polar co-ordinate, and go from there: $$x = r \cos \alpha \quad \mbox{and} \quad y = r \sin \alpha$$
Now (x, y) -> (0, 0) means that r -> 0, and $$\alpha$$ can take whatever value. So if you can show that when r -> 0, the expression will tend to some value independent of $$\alpha$$, then the limit exists there.
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Example:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 2 + y ^ 2}{xy}$$
Change it to polar form, we have:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 2 + y ^ 2}{xy} = \lim_{r \rightarrow 0} \frac{r ^ 2}{r ^ 2 \sin \alpha \cos \alpha} = \lim_{r \rightarrow 0} \frac{1}{\cos \alpha \sin \alpha}$$
That means the limit of that expression does depend on $$\alpha$$, hence the limit does not exist at (0, 0).
Can you get this? :)

3. Mar 4, 2006

### moose

The way I would think about this one is that if x and y are approaching the same thing, why not set y equal to x? That way you would have x^3/2x^2
Now you can think about which one is changing faster, etc.

4. Mar 5, 2006

### pivoxa15

I see. It is clever. Is it the standard way of evaluating multivariable limits? What other ways are there?

5. Mar 5, 2006

### pivoxa15

Your method does not work in most cases.

6. Mar 5, 2006

### devious_

Other ways include approaching the point from various lines, e.g. if you're looking at (x,y) -> (0,0) then sometimes letting (x,y) approach the origin from the x-axis or the y-axis ((x,0) or (y,0)) can help you prove the limit doesn't exist.

7. Mar 5, 2006

### benorin

So we may, upon trying the limit along various curves such as y=x, y=x^2, the x-axis (e.g. y=0), the y-axis (e.g. x=0) which all give the value of the limit to be 0, conjecture the value of the limit to be 0. Now we must prove it:

To prove that $$\lim_{(x,y)\rightarrow (0,0)} \frac{yx ^ 2}{x^2 + y^2}=0,$$ we require that

$$\forall\epsilon >0, \exists \delta >0 \mbox{ such that } 0<\sqrt{x^2 + y^2}<\delta\Rightarrow \left| \frac{yx ^ 2}{x^2 + y^2}-0\right| < \epsilon$$​

(and in case you didn't know, the symbol $$\forall$$ is read "for every", the symbol $$\exists$$ is read "there exists", and the symbol $$\Rightarrow$$ is read "implies".)

The proof is this: We will need the following inequality

$$0\leq y^2\Rightarrow x^2\leq x^2+y^2\Rightarrow \frac{x^2}{x^2+y^2}\leq 1.$$​

Let us work the absolute value term contained in the $$\epsilon ,\delta$$ definition of the limit described above, we have

$$\left| \frac{yx ^ 2}{x^2 + y^2}-0\right| =\frac{|y|x ^ 2}{x^2 + y^2} =|y|\frac{x ^ 2}{x^2 + y^2} \leq |y| = \sqrt{y^2} \leq \sqrt{x^2+y^2} <\delta$$​

we want to choose $$\delta$$ so that the absolute value term we just worked with is always less than $$\epsilon$$ given that $$0<\sqrt{x^2+y^2} <\delta .$$ Evidently, we may choose $$\delta =\epsilon$$ to this end, and the proof is complete upon stating this formally.

I know that PF does not permit full solutions to be posted, however it is understood that proofs of this sort are quite tricky to construct, examples are few in texts that require them, and I will have rendered sufficient pedagogical substance by even successfully relating it.

--Ben

EDIT: Thanks VietDao29, all that short-hand is Greek to me.

Last edited: Mar 5, 2006
8. Mar 5, 2006

### VietDao29

One more way is to use the inequality:
$$x ^ 2 + y ^ 2 \geq 2|xy|$$, one can prove it by doing a little rearrangement, and noticing the fact that: (|x| + |y|)2 >= 0.
I'll give one example that's similar to your problem.
----------------
Example:
Find:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2}$$
Now using the inequality above, we have:
$$\left| \frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2} \right| \leq \left| \frac{x ^ 3 y ^ 3}{2xy} \right| = \frac{1}{2} |x ^ 2 y ^ 2| = \frac{1}{2} x ^ 2 y ^ 2$$
Now as (x, y) tends to (0, 0),
$$\frac{1}{2} x ^ 2 y ^ 2 \rightarrow 0$$, hence $$\left| \frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2} \right| \rightarrow 0$$, thus $$\frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2} \rightarrow 0$$, so we can conclude that:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 3 y ^ 3}{x ^ 2 + y ^ 2} = 0$$
Can you get this? :)
--------------
EDIT:
By the way, benorin:
That symbols is read Delta, the symbol that is read "there exists" should be $$\exists$$ .

Last edited: Mar 5, 2006
9. Mar 5, 2006

### pivoxa15

The method you used in your example looked a bit suspect.

If we follow your way and do the limit(x->0, y->0) of (xy)/(x^2+y^2)
we get the limit being smaller than 1/2 which is wrong since it should have no limit at all.

10. Mar 5, 2006

### Benny

Sorry to intrude on your topic but I would like to know if using polar coordinates is a valid method when the limit is (x,y) -> (0,0). Yes, the angle is arbitrary but each combination of a (r,angle) as r approaches zero corresponds to taking the limit along the straight line. Ie. not every single path is considered.

I can see that converting to polar coordinates is sufficient to verify that the limit does not exist. But is it sufficient to show that a limit exists? I apologise if I missed anything.

11. Mar 5, 2006

### HallsofIvy

Staff Emeritus
No, that's perfectly valid. You do not have to assume the angle is constant as r goes to 0. The point is that, in polar coordinates, the distance from the origin is measured by a single variable, r. r is completely independent of the angle. That was the point VietDao29 made in his first post- the limit, as r goes to 0, depended upon the angle and so the limit itself does not exist.

In your problem, $\frac{xy^2}{x^2+ y^2}$, changing to polar coordinates gives $\frac{r^3cos(\theta)sin^2(\theta)}{r^2}= r cos(\theta)sin^2(\theta)$. What does that go to as r goes to 0? Does it depend on $\theta$?

12. Mar 5, 2006

### Benny

Ok, that's good. Thanks for the clarification.

13. Mar 5, 2006

### VietDao29

Yes, the limit of:
$$\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{xy}{x ^ 2 + y ^ 2}$$ does not exist at (0, 0).
So by using the inequality x2 + y2 >= 2|xy|. We have:
$$\left| \frac{xy}{x ^ 2 + y ^ 2} \right| \leq \frac{1}{2} \left| \frac{xy}{xy} \right| = \frac{1}{2}$$
So that means:
$$-\frac{1}{2} \leq \frac{xy}{x ^ 2 + y ^ 2} \leq \frac{1}{2} \quad (*)$$. Hence, no conclusion can be drawn by looking at the inequality (*). Why? Because we don't know if the expression does converge to some value between -1 / 2, and 1 / 2; or it just diverges.
It's like the limit: $$\lim_{x \rightarrow \infty} \sin x$$ does not exist, although we know that: -1 <= sin(x) <= 1. It's because sin(x) does not tend to any specific number (i.e converges to any value) as x tends to infinity.
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However, if we have $$\lim_{x \rightarrow \alpha} |f(x)| = 0$$, then we also have: $$\lim_{x \rightarrow \alpha} f(x) = 0$$. Why?
It's because $$\forall \varepsilon > 0, \exists \delta > 0 : 0 < |x - \alpha| < \delta \Rightarrow ||f(x)| - 0| = |f(x)| = |f(x) - 0| < \varepsilon$$. Now according to the definition of limit, we also have:
$$\lim_{x \rightarrow \alpha} f(x) = 0$$. Can you get this?
You can do the same to prove that if:
$$\lim_{\substack{x \rightarrow \alpha \\ y \rightarrow \beta}} |f(x, y)| = 0$$ then $$\lim_{\substack{x \rightarrow \alpha \\ y \rightarrow \beta}} f(x, y) = 0$$.
Can you get this? :)
---------------
Now just read my last post again to see if you can understand it.
:)

14. Mar 5, 2006

### benorin

Having verified it, I submit in passing that

$$-\frac{1}{2} \leq \frac{xy}{x ^ 2 + y ^ 2} \leq \frac{1}{2}$$

are the best bounds possible.

15. Mar 5, 2006

### pivoxa15

I see your point. Using polar coords to find limits of multivariable functions is better than your first principles method.

Where did you learn evaluating limits of multivarialbe functions using polar coords? I did not see them in calculus textbooks.

16. Mar 6, 2006

### VietDao29

Yes, using polar co-ordinate is far better, and easier than the second way. :)
And guess what? I learnt how to do it at this site, too... :tongue:

17. Mar 6, 2006

### HallsofIvy

Staff Emeritus
Yes, but that tells you nothing about whether the limit at (0,0) exists.

18. Mar 6, 2006

### benorin

Like I said, "in passing".

19. Mar 7, 2006

### pivoxa15

Are the functions we have been discussing 3D or 2D?

20. Mar 7, 2006

### benorin

Recall that z=f(x,y) is a surface, and hence 3-D.