Limit

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.

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 --> 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.

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?)

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 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.

Ok. If a function of L goes to lunch, it pays. Therefore it is broke!😀

How'd I do?