# Thread: Limit of a 2 variables function

1. ## Limit of a 2 variables function

Say whether or not the limit of $\displaystyle f(x,y)$ when $\displaystyle (x,y)$ tends to $\displaystyle (0,0)$ exists.
$\displaystyle f(x,y)=\frac{xy}{x^2-y^2}-2y$.
My attempt : $\displaystyle \lim _{(x,y) \to (0,0)}f(x,y)=\lim _{x\to 0} f(x,0)=0$.

$\displaystyle \lim _{(x,y) \to (0,0)}f(x,y)=\lim _{x\to 0} f(x,x^2)=-\infty$. As both limits are different, the limit doesn't exist.
However I'm unsure I didn't make an error by taking $\displaystyle y=x^2$ for the second limit. Since when $\displaystyle x=1$, this doesn't even make sense to take such an $\displaystyle y$.
The domain of $\displaystyle f$ is any $\displaystyle (x,y) \in \mathbb{R}^2$ such that $\displaystyle x\neq \pm y$.

2. Originally Posted by arbolis
Say whether or not the limit of $\displaystyle f(x,y)$ when $\displaystyle (x,y)$ tends to $\displaystyle (0,0)$ exists.
$\displaystyle f(x,y)=\frac{xy}{x^2-y^2}-2y$.
My attempt : $\displaystyle \lim _{(x,y) \to (0,0)}f(x,y)=\lim _{x\to 0} f(x,0)=0$.

$\displaystyle \color{red}\lim _{(x,y) \to (0,0)}f(x,y)=\lim _{x\to 0} f(x,x^2)=-\infty$.
i don't think that's true. i get that the limit is also 0 along y = x^2

As both limits are different, the limit doesn't exist.
However I'm unsure I didn't make an error by taking $\displaystyle y=x^2$ for the second limit. Since when $\displaystyle x=1$, this doesn't even make sense to take such an $\displaystyle y$.
The domain of $\displaystyle f$ is any $\displaystyle (x,y) \in \mathbb{R}^2$ such that $\displaystyle x\neq \pm y$.
your method is sound though. note that if we approach (0,0) along x = 2y we get -2/3 as the limit. this means the limit doesn't exist

(the curve you approach the origin must actually go through the origin. the line x = 1 does not.)

3. Originally Posted by arbolis
Say whether or not the limit of $\displaystyle f(x,y)$ when $\displaystyle (x,y)$ tends to $\displaystyle (0,0)$ exists.
$\displaystyle f(x,y)=\frac{xy}{x^2-y^2}-2y$.
My attempt : $\displaystyle \lim _{(x,y) \to (0,0)}f(x,y)=\lim _{x\to 0} f(x,0)=0$.

$\displaystyle \lim _{(x,y) \to (0,0)}f(x,y)=\lim _{x\to 0} f(x,x^2)=-\infty$. As both limits are different, the limit doesn't exist.
However I'm unsure I didn't make an error by taking $\displaystyle y=x^2$ for the second limit. Since when $\displaystyle x=1$, this doesn't even make sense to take such an $\displaystyle y$.
The domain of $\displaystyle f$ is any $\displaystyle (x,y) \in \mathbb{R}^2$ such that $\displaystyle x\neq \pm y$.
If the given function approaches (0, 0) along the line y = mx ($\displaystyle m \neq \pm 1$), the limiting value is $\displaystyle \frac{m}{1 - m^2}$. So clearly the limiting value depends on how you approach (0, 0) ....

4. Originally Posted by Jhevon
i don't think that's true. i get that the limit is also 0 along y = x^2
I get it now, thanks.
Originally Posted by mr fantastic
If the given function approaches (0, 0) along the line y = mx (), the limiting value is . So clearly the limiting value depends on how you approach (0, 0) ....
I wasn't aware of this! I mean I didn't realize it. This holds for this particular example, right? If it hold for all example then no 2 variable functions would have limit in (0,0), right? I need to clear this doubt!

5. Originally Posted by arbolis
This holds for this particular example, right? If it hold for all example then no 2 variable functions would have limit in (0,0), right? I need to clear this doubt!
For this particular problem. Not all.

6. Originally Posted by Plato
For this particular problem. Not all.
Thanks, makes sense. I was a bit scared for a moment.

7. There are, after all, some limits that really exist at (0,0).

8. Originally Posted by HallsofIvy
There are, after all, some limits that really exist at (0,0).
Ahah yes.