- Thread starter
- #1
GreenGoblin
Member
- Feb 22, 2012
- 68

i used x in the sense that $x \in \mathbb{R} \times \mathbb{R}$, so x = ($x_1,x_2$)huh?i posted a pica. of the problem. that cant be the answer because this is not x, this is (x,y) so the answer has to be as much to do with y as x.
what has that got to do with anything? i already dont like you.Take the following function
$$f(x,y) = \left\{
\begin{array}{lr}
(x+y+1,\frac{x^2-y^2}{x^2+y^2}) & : (x,y) \not = (0,0)\\
(1,0) & : (x,y) = (0,0)
\end{array}
\right.
$$
Then f(0,0) = (1,0)
But Take the sequence $X_j$ = ($\frac{1}{j}$,0) where $j \in \mathbb{N}$
then $X_j \to (0, 0)$ as $j \to \infty$
but $lim_{j \to \infty}f(0,\frac{1}{j}) = (\frac{1}{j} + 1, 1) \to (1,1)$
but (1,1) $\not = f(0,0) = (1,0)$.
Those were the magic words. Thread closed.what has that got to do with anything? i already dont like you.