Limit of a two variable function

In summary, the conversation discusses verifying the limit of a function as (x,y) approaches (0,0), and explores different methods to solve it. The conversation also mentions using the L'Hopital's rule and the epsilon-delta method to prove the limit. The limit is eventually found to be 0 after separating the function into two parts.
  • #36
See my note
 
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  • #37
Stephen Tashi said:
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 [itex]\sin (\theta) ~ \theta[/itex] for small [itex]\theta[/itex]) 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:

[tex]\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[/tex]

So the original limit does not exist.
 
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Likes Stephen Tashi and Inventive
  • #38
Citan Uzuki said:
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 [itex]\sin (\theta) ~ \theta[/itex] for small [itex]\theta[/itex]) 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:

[tex]\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[/tex]

So the original limit does not exist.
Thanks for taking the time to explain it this way. I can visualize what you are saying
 
  • #39
Citan Uzuki said:
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##.
 
  • #40
Thank you Stephen for the further information
 
  • #41
<h2>1. What is the definition of a limit of a two variable function?</h2><p>The limit of a two variable function is the value that the function approaches as the two variables approach a particular point or value. It is also known as the "limit point" or "limiting value".</p><h2>2. How is the limit of a two variable function calculated?</h2><p>The limit of a two variable function is calculated by evaluating the function at different points that approach the given point, and observing the trend of the values. If the values of the function approach a single value as the points get closer to the given point, then that value is considered as the limit.</p><h2>3. What is the significance of the limit of a two variable function?</h2><p>The limit of a two variable function helps us understand the behavior of the function at a particular point. It also helps in determining the continuity of the function at that point, and in finding the derivatives of the function.</p><h2>4. Can the limit of a two variable function have multiple values?</h2><p>No, the limit of a two variable function can only have one value. This value may be the same or different from the value of the function at the given point.</p><h2>5. How is the limit of a two variable function affected by the direction of approach?</h2><p>The limit of a two variable function may be different for different directions of approach. This is known as the directional limit. It is important to consider the direction of approach when calculating the limit of a two variable function.</p>

1. What is the definition of a limit of a two variable function?

The limit of a two variable function is the value that the function approaches as the two variables approach a particular point or value. It is also known as the "limit point" or "limiting value".

2. How is the limit of a two variable function calculated?

The limit of a two variable function is calculated by evaluating the function at different points that approach the given point, and observing the trend of the values. If the values of the function approach a single value as the points get closer to the given point, then that value is considered as the limit.

3. What is the significance of the limit of a two variable function?

The limit of a two variable function helps us understand the behavior of the function at a particular point. It also helps in determining the continuity of the function at that point, and in finding the derivatives of the function.

4. Can the limit of a two variable function have multiple values?

No, the limit of a two variable function can only have one value. This value may be the same or different from the value of the function at the given point.

5. How is the limit of a two variable function affected by the direction of approach?

The limit of a two variable function may be different for different directions of approach. This is known as the directional limit. It is important to consider the direction of approach when calculating the limit of a two variable function.

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