I'm having a little trouble using delta and epsilon to prove the following limit :
$\lim\limits_{(x,y) \to (1,0)} \frac{xy-y}{(x-1)^2 + y^2}$
So I hypothesize that the limit is zero, based on approaching the point (1,0) from the lines $y = 0$, $x = 1$, and $y = mx - m$. However, when I try to give a rigorous proof using delta and epsilon, I get stuck.
I set up the problem as :
if $0 < \sqrt{(x-1)^2 + y^2} < \delta$ then $\lvert\frac{xy-y}{(x-1)^2 + y^2} - 0\rvert < \epsilon$
Simple enough. But when I try to simplify the function, I can't seem to make any headway trying to manipulate it into something that I can directly relate to delta.
I basically can rewrite the function as $\frac{\lvert x-1\rvert \lvert y\rvert}{(x-1)^2 + y^2}$, but I'm having trouble finding a way to draw any conclusions. I end up getting that the whole equation is less than one, but that doesn't seem to help.
Any insight will be immensely appreciated ! (BTW, this is not just a HW question, I really want to understand what's going on in this limit, simple as it may be.)
The limit doesn't exist.
If you approach $(1,0)$ along the line $y=0$, the limit is $0$.
If you approach $(1,0)$ along the line $y=x-1$, the limit is ${\large{\frac{1}{2}}}$.
To see it more easily, note that $$\frac{xy-y}{(x-1)^2+y^2}=\frac{y(x-1)}{(x-1)^2+y^2}$$ hence, replacing $x-1$ by $u$ and $y$ by $v$, the limit can be recast as $$\lim_{(u,v)\to (0,0)}\frac{uv}{u^2+v^2}$$ and then, approaching $(0,0)$ along the lines $u=0$ and $u=v$, you get different limits.