28 Jan '04 04:321 edit

Let f:R^2 --> R be such that for all positive integers N, f(1/N,0) = 1 and f(0,1/N) = -1. Does there exist real L such that f(x,y) --> L as (x,y) --> (0,0)? Prove it.

- Joined
- 29 Nov '02
- Moves
- 17317

Elsewhere- Joined
- 18 Aug '03
- Moves
- 54533

Chesstralia28 Jan '04 05:071 edit

you don't seem to have defined f off the axes - why is this?*Originally posted by royalchicken***Let f:R^2 --> R be such that for all positive integers N, f(1/N,0) = 1 and f(0,1/N) = -1. Does there exist real L such that f(x,y) --> L as (x,y) --> (0,0)? Prove it.**

as for proof: consider integer N --> big positive.

then f(0,1/N) --> -1

so L must be -1 (if such L exists)

also f(1/N,0) --> 1

so L = 1, contradiction

so no such L,

so f is naughty at (0,0)

(does this have a bigger significance?)

edit for non-maths folks: you get a topographical map. there is a place on this particular map where as you walk towards it from one direction you go up high in the clouds, if you walk towards that same place from a different direction you end up under the sea. a bit like a cliff edge.- Joined
- 29 Nov '02
- Moves
- 17317

Elsewhere28 Jan '04 05:09

It doesn't need to be defined off the axes--it is not completely defined ON the axes, but you are correct.*Originally posted by flexmore***you don't seem to have defined f off the axes - why is this?**

as for proof: consider integer N --> big positive.

then f(0,1/N) --> -1

so L must be -1 (if such L exists)

also f(1/N,0) --> 1

so L = 1, contradiction

so no such L,

so f is naughty at (0,0)

(does this have a bigger significance?)- Joined
- 26 Jan '03
- Moves
- 1644

x10,y45,z-88,t3.141531 Jan '04 21:56

Ok. If a function of L goes to lunch, it pays. Therefore it is broke!ðŸ˜€*Originally posted by royalchicken***Let f:R^2 --> R be such that for all positive integers N, f(1/N,0) = 1 and f(0,1/N) = -1. Does there exist real L such that f(x,y) --> L as (x,y) --> (0,0)? Prove it.**

How'd I do?