This implies that
Now just let and choose ...
Consider R^2 where p((a,b),(c,d))=Sqrt((a-c)^2+(b-d)^2) and R where p(x,y) = |x-y|. f:R^2->R f((x,y))=x+y. Prove Lim (x,y)->(1,2) (x + y) = 3.
I really don't know quite where to start this problem. I know the limit will exist, if it exists, in R. I think I might need to know the distance from (x,y) to (1,2). But then where would I go with that? Thanks for your help.
Here's a rather different way to do it: Let u= x- 1, v= y- 2. Then x= u+ 1, y= v+ 2 so x+ y= u+ v+ 3 and the problem becomes .
Now convert to polar coordinates: , so that . The point of that is that now the single variable "r" measures the distance from (u, v) to (0, 0). For any , and are both less than or equal to 1 so is less than or equal to 2. .