# 0/0 as the result of taking a limit

1. Dec 17, 2004

### quasar987

I know when we encounter 0/0 as the result of taking a limit, we cannot conclude to the value of the limit. But what about 0/0 for a function? I am confused because in last year's final calculus exam, there's a question that reads:

Calculate the gradient vector of the function

$$f(x,y)=\frac{x^4+y^3}{x^2+y^2}$$

at all point (including at (0,0)).

But for exemple,

$$\left[ \frac{\partial f}{\partial x} \right]_{(0,0)} = \left[\frac{4x^3}{x^2+y^2}-\frac{2x^5}{(x^2+y^2)^2}-\frac{2xy^3}{(x^2+y^2)^2}\right]_{(0,0)} = \frac{0}{0}$$

2. Dec 17, 2004

### dextercioby

You missunderstood a great deal of mathematics.When u're given a function,u have to see first at which points (what set) it is defined.In your case,neither the function,nor the derivatives are defined in (0,0),so your calculation is pointless.I should say INCORRECT.

Daniel.

PS.Was it written on the exam paper "including (0,0)"??If so,it must have been a (bad) joke from the teacher.

3. Dec 17, 2004

### quasar987

Yes, I transcripted the question as it was written on the paper!

And the worst part is that a friend of mine went to see a teacher about that question today and he told him the way to do it was to use the limit to show that the gradient was (0,0) at (0,0) !! Insane! :yuck:

The exam is at: http://www.dms.umontreal.ca/~giroux/documentsc1/finalC1H04.html and it is the first question. Although it is in french, you can see that the "including at (0,0)" part is there.

Last edited: Dec 17, 2004
4. Dec 17, 2004

### dextercioby

Il n'y a acune problème.Je comprends Français.
Yes,i saw that.He meant to 'compute that limit'.It's (0,0),as it can easily be shown.However,it is still weird.He could have said:"Find the limit."Students would have gotten it.

Daniel.

PS.Je pense que le proffesseur Giroux (?) est un idiot. :tongue2:

5. Dec 18, 2004

### Integral

Staff Emeritus
Oh My, Oh My!

you must keep in mind that this is function of 2 variables, You need to be a bit more careful.

First note that if you plot points on either side of zero the function is well behaved.

The correct way to analysis multi variable limits is to hold one variable constant, while the limit in the other variable is examined.

so if you first evaluate your partial of f wrt x at y=0, you have

$$\left[ \frac{\partial f}{\partial x} \right]_{(x,0)} = \left[\frac{4x^3}{x^2}-\frac{2x^5}{(x^2)^2}-\frac{0}{(x^2)^2}\right]_{(x,0)}=2x$$

Clearly this is 0 at x = 0

You will get similar results if you hold x constant (that is compute the partial wrt y) and take the y limit as x is held constant.

You will find that the function is well behaved and approaching zero for all x,y near zero. therefore the limit exists at (0.0) and is 0. This can also be seen by plotting points near zero.

6. Dec 18, 2004

### quasar987

But do we agree that finding the limit of the gradient as (x,y) approches (0,0) is not the same as "calculating" the gradient for the value (0,0) ?

Are you saying it is not necessary to use the epsilon-delta definition to show that the limit is 0?

7. Dec 18, 2004

### arildno

1. Prove that
$$\lim_{(x,y)\to(0,0)}f(x,y)=0$$
We must show that given $$\epsilon>0$$ there exist a $$\delta>0$$
so that $$||\vec{x}-\vec{0}||<\delta\to|f(\vec{x})-0|<\epsilon, \vec{x}\equiv(x,y)$$
Rewrite your function in polar form:
$$x=r\cos\theta,y=r\sin\theta\to{f}(x,y)=g(r,\theta)=r^{2}\cos^{4}\theta+r\sin^{3}\theta$$
Hence, since:
$$|g(r,\theta)-0|<=r^{2}+r$$
it follows that we can find, for $$\epsilon>0$$ a $$\delta>0$$ so that
$$r<\delta\to{r}^{2}+r<\epsilon$$
$$\delta<min(\sqrt{\frac{\epsilon}{2}},\frac{\epsilon}{2})$$
is perhaps the simplest choice.

Hence, defining f(0,0)=0 gives us a continous function.

8. Dec 18, 2004

### arildno

2. By definition,
$$\frac{\partial{f}}{\partial{x}}\mid_{(x_{0},y_{0})}=\lim_{x\to{x}_{0}}\frac{f(x,y_{0})-f(x_{0},y_{0})}{x-x_{0}}$$
$$f(x,0)=x^{2},f(0,0)=0,f(0,y)=y$$
We therefore gain:
$$\frac{\partial{f}}{\partial{x}}\mid_{0,0}=\lim_{x\to0}x=0,\frac{\partial{f}}{\partial{y}}\mid_{0,0}=\lim_{y\to0}1=1$$

Hmm..I'll go and check up the definition to be sure, but I'm pretty certain I'm right here..

Last edited: Dec 18, 2004
9. Dec 18, 2004

### arildno

3. It is NOT correct that, for example $$\frac{\partial{f}}{\partial{x}}$$ is continuous at (0,0):
We compute $$\frac{\partial{f}}{\partial{x}}$$ at $$(x_{0},y_{0})\neq(0,0)$$
$$\frac{\partial{f}}{\partial{x}}\mid_{(x_{0},y_{0})}=\frac{4x_{0}^{3}(x_{0}^{2}+y_{0}^{2})-2x_{0}(x_{0}^{4}+y_{0}^{3})}{(x_{0}^{2}+y_{0}^{2})^{2}}$$

Rewriting in polar form:
$$\frac{\partial{f}}{\partial{x}}=h(r,\theta),x_{0}=r\cos\theta,y_{0}=r\sin\theta$$
We get:
$$h(r,\theta)=r(4\cos^{3}\theta-2\cos^{5}\theta)-2\cos\theta\sin^{3}\theta$$

Hence,
$$\lim_{(x_{0},y_{0}\to(0,0)}\frac{\partial{f}}{\partial{x}}\mid_{(x_{0},y_{0})}$$
depends on the angle by which we approach the origin.

In particular, the limit value will not always equal zero
,which is the function value $$\frac{\partial{f}}{\partial{x}}\mid_{(0,0)}$$
Hence the x-derivative is discontinuous at the origin.

Last edited: Dec 18, 2004
10. Dec 18, 2004

### arildno

4. It is instructive to compute the DIRECTIONAL DERIVATIVES at the origin.
by definition, the derivative in the direction $$\vec{n},||\vec{n}||=1$$
is given by:
$$\frac{\partial{f}}{\partial{n}}\mid_{(x_{0},y_{0})}\equiv\lim_{t\to0}\frac{f(x_{0}+tn_{x},y_{0}+tn_{y})-f(x_{0},y_{0})}{t}$$
Setting $$\vec{n}=n_{x}\vec{i}+n_{y}\vec{j}=\cos\theta\vec{i}+\sin\theta\vec{j}$$
we gain:
$$\frac{\partial{f}}{\partial{n}}\mid_{(0,0)}=\sin^{3}\theta$$

11. Dec 18, 2004

### Sirus

Hmm...mais vous n'écrivez pas si bien...

I kidd.

12. Dec 19, 2004

### dextercioby

Oui,c'est vrai.J'ai oublié un "u"... "aucune",peut $\hat{e}$tre...??

Daniel.

13. Dec 19, 2004

### quasar987

Thanks arildno. I don't think this is what the question asked but they were assuredly very instructive posts, especially #4.

14. Dec 19, 2004

### arildno

Just a final comment.
This function is an example where the partial derivatives all exist at (0,0), but still is not a differentiable function at (0,0).
If you want "gradient" to mean the "(total) derivative of f", then the gradient
doesn't exist at (0,0).

15. Dec 19, 2004

### mathwonk

By definition, the total derivative of f(x,y) equals 0 at (0,0) if and only if

f(x,y)/|(x,y)| goes to zero as (x,y) does, where |(x,y)| = sqrt(x^2+y^2). in the case above f(x,y)/|(x,y)| = {x^4 + y^3}/(x^2+y^2)^(3/2).

thus the bottom has homogeneous degree 3, while the top has partly degree 4 and partly degree 3. thus if you approach (0,0) along the line x=0, this quotient is equal to 1, hence does not approach 0.

can you see how to modify the top of f slightly so as to get a function which does have total derivative zero at (0,0)? and by the way the professor is not an idiot here. indeed this is the kind of question which exposes whether one does or does not know what a derivative means.

16. Dec 20, 2004

### arildno

In order to prove that f is not differentiable at zero (that is, that the "total derivative" doesn't exist), you must show that there cannot exist a vector (a,b) independent of $$\vec{x}=(x,y)$$ so that:
$$\lim_{(\vec{x})\to(0,0)}\frac{|f(x,y)-f(0,0)-ax-by|}{||\vec{x}||}=0$$