# As (a,b)->(0,0) limit (a,b)/(a^2+b^2) exists

1. Sep 21, 2006

### precondition

As (a,b)-->(0,0) limit (a,b)/(a^2+b^2) exists

I'm stuck on epsilon delta argument for (a,b)-->(0,0) limit (ab)/(a^2+b^2) exists. HELP

Last edited: Sep 21, 2006
2. Sep 21, 2006

### HallsofIvy

Staff Emeritus
For problems like this, I recommend converting to polar coordinates:

$a= r cos(\theta) b= r sin(\theta)$ so
$$\frac{ab}{a^2+ b^2}= \frac{r^2 sin(\theta)cos(\theta)}{r^2}$$

$$= cos(\theta)sin(\theta)$$
The distance from (0,0) to (a,b) is just r.
( Hmm, I forsee a serious problem in proving that limit exists!)

In the title you said (a,b)/(a^2+ b^2) which I would interpret as the vetor
$$(\frac{a}{a^2+ b^2},\frac{b}{a^2+b^2})$$
In polar coordinates that is
$$(\frac{cos(\theta)}{r}, \frac{sin(\theta)}{r})$$
Now it is easy to see that each component is less than 1/r and, again, r measures the distance from (0,0) to the point (a,b).

3. Sep 21, 2006

### precondition

umm... again sorry but
function is (ab)/(a^2+b^2) comma shouldn't be there...
sorry

4. Sep 22, 2006

### istevenson

Yes, you have seen it. actually the limit doesnt exist,because the parameter $$\theta$$ in your equation isnt a constant.

5. Sep 22, 2006

### HallsofIvy

Staff Emeritus
Well, in that case, there is no "epsilon-delta" argument that the limit exists for a very good reason. The limit does not exist.