Limit of a two variable function

[Inventive]

See my note

 

[Citan Uzuki]

That's correct, but does that, by itself, prevent the limit from existing? This is very technical question involving how we interpret the definition of the limit. The definition has the form " For each ... there exists...such that...for each $(x,y)$, $| f(x,y) - L| < \epsilon$". $\$ Are we to understand that the implicit meaning is "for each $(x,y)$ in the domain of $f$"? $\$ Or perhaps that $| f(x,y) - L | < \epsilon$" is not a false statement when $(x,y)$ is not in the domain of $f$, but rather an undefined expression?

All very true, but actually orthogonal to the point I was trying to make. What I was trying to get at is that if you are trying to find the limit of a rational function (or something which is effectively a rational function, since $\sin (\theta) ~ \theta$ for small $\theta$) at the origin, the denominator being zero along some curve running through the origin usually means that you can make the function blow up by following another curve close to the first one. Thus seeing that the function fails to exist along some curve through the origin is usually a good sign that the limit won't exist, independently of the answer to that technical question.

That's a good idea, but does it work? $\ | sin(\theta) | \le |\theta |$. $\ \theta$ would be a polynomial of degree 12 in $y$ and the denominator would be a polynomial of degree 4 in $y$.

Yes. Along the curve x=-y^2 + y^4:

$$\lim_{y \to 0^+} \frac{\sin (x^3 + y^3)}{x + y^2} \\ = \lim_{y \to 0^+}\frac{\sin ((-y^2 + y^4)^3 + y^3)}{-y^2 + y^4 + y^2} \\ = \lim_{y \to 0^+}\frac{\sin (y^3 - y^6 + 3y^8 - 3y^{10} + y^{12})}{y^4} \\ = \lim_{y \to 0^+}\frac{y^3 + O(y^6)}{y^4} \\ = \lim_{y \to 0^+}\frac{1 + O(y^3)}{y} \\ = \infty$$

So the original limit does not exist.

 

[Inventive]

All very true, but actually orthogonal to the point I was trying to make. What I was trying to get at is that if you are trying to find the limit of a rational function (or something which is effectively a rational function, since $\sin (\theta) ~ \theta$ for small $\theta$) at the origin, the denominator being zero along some curve running through the origin usually means that you can make the function blow up by following another curve close to the first one. Thus seeing that the function fails to exist along some curve through the origin is usually a good sign that the limit won't exist, independently of the answer to that technical question.
Yes. Along the curve x=-y^2 + y^4:

$$\lim_{y \to 0^+} \frac{\sin (x^3 + y^3)}{x + y^2} \\ = \lim_{y \to 0^+}\frac{\sin ((-y^2 + y^4)^3 + y^3)}{-y^2 + y^4 + y^2} \\ = \lim_{y \to 0^+}\frac{\sin (y^3 - y^6 + 3y^8 - 3y^{10} + y^{12})}{y^4} \\ = \lim_{y \to 0^+}\frac{y^3 + O(y^6)}{y^4} \\ = \lim_{y \to 0^+}\frac{1 + O(y^3)}{y} \\ = \infty$$

So the original limit does not exist.

Thanks for taking the time to explain it this way. I can visualize what you are saying

 

[Stephen Tashi]

Yes. Along the curve x=-y^2 + y^4:

I see what you mean.

We can also use the way more often seen in textbooks - showing that different limits can be had by using different curves if we use curves of the form $x = -y^2 + ky^3$.

Let $G(x,y) = \frac{x^3 + y^3}{x + y^2}\$ Use the path defined by $x = -y^2 + k y^3$ where $k$ is a non-zero constant. This path goes through $(0,0)$
Along this path, for $y \ne 0$, $G(x,y) = g(y) = \frac{ (-y^2 + ky^3)^3 + y^3}{ky^3} = \frac{(y^6 + 3ky^7 - 3k^2 y^8 + k^3 y^9 ) + y^3}{ky^3} = \frac{y^3+ 3ky^4 - 3k^2y^5 + y^6 + 1}{k}$
So $lim_{y \rightarrow 0} g(y) = \frac{1}{k}$.
Since $\sin()$ is a continuous function $lim_{y \rightarrow 0} \sin(g(y)) = \sin( \frac{1}{k})$.
This shows $lim_{(x,y) \rightarrow (0,0)} sin(G(x,y))$ does not exist since we have can get different limits on different paths depending on our choice for $k$.

 

[Inventive]

Thank you Stephen for the further information

 

[haushofer]

That is not a valid conclusion. The existence of a common value for the limit along a family of straight line paths does not imply that the limit as (x,y) -> (0,0) exists. For example, see the answer to http://math.stackexchange.com/quest...existence-of-limit-of-a-two-variable-function

You're right. The first example of your link is even used in stewart's textbook. Thanks for noting!