Limit of a two variable function

[Stephen Tashi]

This limit does not exist. Note that the function isn't even defined on the path x=-y^2,

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?

and if you consider paths close to that path (e.g. x = -y^2 + y^4), the limit diverges to infinity.

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 $y$.

 

[Stephen Tashi]

I'm trying to verify that: $$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x+y^{2}}=0.$$

One thought is that $|\sin(\theta)| \le |\theta|$. $\$ Argue that for $(x,y) \ne (0,0)$, $$ | \frac{ \sin(x^3 + y^3)}{x + y^2} | \le |\frac{x^3 + y^3}{x + y^2}|$$. Then try polar coordinates on the ratio of the polynomials.

$$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x^{3}+y^{3}}\frac{x^{3}+y^{3}}{x+y^{2}}=\lim_{(x,y)\rightarrow (0,0)} \frac{x^{3}+y^{3}}{x+y^2}$$

That's a good intuitive idea, but notice how hard it is to make that algebra into correct logic. For example, it assumes at the outset that $lim_{(x,y)\rightarrow (0,0)}\frac{\sin (x^{3}+y^{3})}{x^{3}+y^{3}}\frac{x^{3}+y^{3}}{x+y^{2}}$ exists. Of course, some instructors (and many home work graders) prefer to see algebra instead of logic.

The basic pattern that obeys logic would be:
Show $lim_{(x,y) \rightarrow (0,0)} p(x,y) = L_1$
Show $lim_{(xy) \rightarrow (0,0)} q(x,y) = L_2$
Conclude that $lim_{(x,y) \rightarrow (0,0)} p(x,y)q(x,y) = L_1 L_2$.

Polar coordinates might be a good way to tackle $\lim_{(x,y)\rightarrow (0,0)} \frac{x^{3}+y^{3}}{x+y^2}$ instead of using further algebra with $x$'s and $y$'s.

 

[Inventive]

How about this approach. I plotted this function using quick graph. There are no breaks in the graph at the origin so therefore it is continuous there. Also the numerator sin(x^3 + y^3) can be rewritten as sin(x^3)cos(y^3) + cos(x^3)sin(y^3). Now that the numerator has been rewritten as a sum of the products of periodic continuous functions it should make it easier to prove the existence of the limit. Even though the denominator is undefined at (0,0), the numerator is approaching 0 faster than the denominator in terms of how the graph is changing.

 

[Stephen Tashi]

There are no breaks in the graph at the origin so therefore it is continuous there.

The function $\frac{\sin(x^3 + y^3)}{x + y^2}$ is not continuous at $(0,0)$. Have you studied the definition of "$f(x,y)$ is continuous at $(a,b)$" ?

 

[Inventive]

i don't understand the "*. *". Comment. This site is not about PROVING someone is wrong or putting people down which you did with me. The dialog in these forums is not supposed to be judgmental towards others. So tell me why you disagree without being judgmental. What do you see in detail is wrong with my approach ?

 

[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:

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

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:

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

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!