 Posers and Puzzles

1. 28 Jan '04 04:321 edit
Let f:R^2 --&gt; 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) --&gt; L as (x,y) --&gt; (0,0)? Prove it.
2. 28 Jan '04 05:071 edit
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.
you don't seem to have defined f off the axes - why is this?

as for proof: consider integer N --&gt; big positive.

then f(0,1/N) --&gt; -1

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

also f(1/N,0) --&gt; 1

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.
3. 28 Jan '04 05:09
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

4. 31 Jan '04 21:56