Finding a limit with two variables

Question

If $(x,y)\neq (0,0)$, let $f(x,y)=(x^2-y^2)/(x^2+y^2)$. Find the limit of f(x,y) as $(x,y) \leftrightarrow (0,0)$ along the line y=mx.

By replacing y=mx, I found $f(x,y)=(1-m^2)/(1+m^2)$

Is it possible to define f(0,0) so as to make f continuous at (0,0)

This I didn't know

Please help

score 1 · Accepted Answer · answered Mar 09 '13 at 23:58

1

If a limit $L$ at $(0,0)$ existed, then for every $m$, you would have $$ \lim_{x\rightarrow 0}f(x,mx)=L $$ since $(x,mx)\rightarrow (0,0)$ when $x\rightarrow 0$.

Given your work, this yields $$ \frac{1-m^2}{1+m^2}=L\qquad\forall m\in\mathbb{R}. $$ Can you find a contradiction? Try two simple values of $m$.

answered Mar 09 '13 at 23:58

Julien

44,791

No I cannot. Can specify a bit more – Mar 10 '13 at 00:00
@Carpediem Ok, I edited. And now? – Julien Mar 10 '13 at 00:02
We get a different limit for different values of m – Mar 10 '13 at 00:05
@Carpediem Yes, e.g. $m=0$ and $m=1$ yield $L=1$ and $L=0$. Contradiction. – Julien Mar 10 '13 at 00:06
So the answer to the second question is no – Mar 10 '13 at 00:07
If there is a continuous extension of $f$ by $f(0,0)=L$ at $(0,0)$, you would have $L=f(0,0)=\lim_{(x,y)\rightarrow (0,0)} f(x,y)$. So the answer is indeed no. – Julien Mar 10 '13 at 00:09
let us continue this discussion in chat – Mar 10 '13 at 00:11

Finding a limit with two variables

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