Originally Posted by

**arbolis** Let

and let

be defined in all

by

.

1)Analyze the continuity of

in all

.

I'll put the other questions if I need help, but I'd like a confirmation of my result for part 1).

My attempt :

First I tried to visualize the function and it was quite hard. I couldn't even do that. I realize that it takes the element

(a line) from the domain and send it to the image as the line

(or curve?). I also realize that it sends all the elements under the curve

into another area, but I didn't visualize it. Anyway, back to the question, I think the limit exists by intuition.

Here's my work to support this :

is continuous in all

except maybe in

.

Right. But you could have been a bit more explicit here. For example you could have argued like this:

A is the half-plane below (and including) the line

. f happens to be continuous at any

*inner* point of either A or

: because f happens to be identical to a continuous function in an entire neighborhood of such a point.

Thus, the only problematic points are those on the line

.

But note: by arguing like this you already assume that f restricted to A happens to be continuous, so that there is no reason to even try to prove it only for points on the line the way you do below:

I'll try to prove it is :

I want to prove that

.

,

such that if

.

which work if I take

. Hmm no it doesn't work. I must sleep now, will think about it.

If you have any help to provide, it's always a pleasure for me. Don't hesitate.

Maybe you find it difficult to prove that goes to as tends to . Initially I wouldn't have thought of even trying to prove this: one usually assumes that a polynomial function of the coordinates of a point of is continuous without further addo. This is because the projection of a point of onto its k-th coordinate is continuous, and that sums and products of continous functions are continuous.

Thus there is no reason to prove continuity of a polynomial in the coordinates for each and every such polynomial using a pure delta-epsilon argument: it would be too much work (depending on the complexity of the polynomial) and it wouldn't be particularly enlightening either.