# Using the Delta-epsilon definition but for two variables?

1. Sep 10, 2007

### pr0me7heu2

I have probably over thought the whole thing... but I can't seem to find any place to start with this one:

Using the formal definition of a limit:

f(x,y)= y / (x^2 + 1) e=0.05

2. Sep 10, 2007

### D H

Staff Emeritus
For multiple dimensions you need to extend the epsilon-delta notation to incorporate the epsilon-neighborhood of the point p0 = (x,y), which is the set of all points p such that the distance between p0 and p is less than epsilon.

3. Sep 10, 2007

### Kummer

The limit seems to be 0.
So if $$0<|y|<\delta$$ and $$0<|x|<\delta$$.

$$\left| \frac{y}{x^2+1} \right| \leq \frac{|y|}{x^2} < \frac{\delta}{x^2}$$.

Now use single variable analysis that $$\frac{\delta}{x^2}$$ can be made sufficiently small.

4. Sep 11, 2007

### HallsofIvy

Staff Emeritus
limit as (x,y) goes to what?