Finding the Limit: (x2y)/(x4 + y2)

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In summary, the limit of (x2y)/(x4 + y2) as (x,y) approaches (0,0) does not exist. Approaching the problem using polar coordinates may lead to incorrect results. It is important to make sure substitutions are made correctly when solving this limit.
  • #1
physics=world
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Homework Statement



Find the limit:

the limit of (x2y)/(x4 + y2) as (x,y) approaches (0,0)


Homework Equations





The Attempt at a Solution



I took the limit of the numerator and denominator separately.

The numerator equals to 0 as well as the denominator.

So, I get the indeterminant form 0/0.

Where do I go from here?
 
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  • #2
Try approaching the problem using polar coordinates:

[itex]x=r\cos{\theta}[/itex]
[itex]y=r\sin{\theta}[/itex]
 
  • #3
electricspit said:
Try approaching the problem using polar coordinates:

[itex]x=r\cos{\theta}[/itex]
[itex]y=r\sin{\theta}[/itex]


After solving I get cos2(theta) / sin(theta)
 
  • #4
You could choose ##\delta = min\{1,2 \epsilon\}##
 
  • #5
Uh according to what I have, I'll do the numerator and let you do the denominator:

[itex]
x^2 y = r^3 \cos{\theta}^2\sin{\theta}
[/itex]

Also remember [itex](x,y)\to 0[/itex] is the same as [itex]r\to 0[/itex]
 
  • #6
Can I choose y = 0 and find the limit. Then, y = x2 and find the limit. Will that show that it approaches different limits and Therefore, it would not exist?
 
  • #7
I get the limit to exist. Make sure you are making the substitutions correctly.
 
  • #8
electricspit said:
Uh according to what I have, I'll do the numerator and let you do the denominator:

[itex]
x^2 y = r^3 \cos{\theta}^2\sin{\theta}
[/itex]

Also remember [itex](x,y)\to 0[/itex] is the same as [itex]r\to 0[/itex]

denominator:

x4 + y2 = r4cos4(theta) + sin2(theta)
 
  • #9
Ah see you are missing the r squared term in front of your sine term.
 
  • #10
electricspit said:
I get the limit to exist. Make sure you are making the substitutions correctly.
I get that the limit doesn't exist.
 
  • #12
Let (x, y) → (0, 0) along either of the paths, y = x2 and y = -x2 .
 
  • #13
Touché, WolframAlpha lies, as my polar coordinate method is not sufficient.
 
  • #14
physics=world said:
After solving I get cos2(theta) / sin(theta)

You shouldn't. You made a mistake.
 
  • #15
electricspit said:
Touché, WolframAlpha lies, as my polar coordinate method is not sufficient.
I stumbled on that path quite by accident, although I did use WolframAlpha to help.

I was somewhat surprised that WolframAlpha messed up evaluating this limit.
 
  • #16
Yeah I just used the polar coordinates method and assumed Wolfram would have given the correct answer.
 

1. What is the limit of (x2y)/(x4 + y2) as x and y approach 0?

The limit of (x2y)/(x4 + y2) as x and y approach 0 is 0. This can be determined by plugging in x = 0 and y = 0 into the equation, which results in 0/0. Using algebraic manipulation and L'Hopital's rule, we can simplify the equation to (2y)/(4x3 + 2y), and then plug in the values of x = 0 and y = 0 to get a final answer of 0.

2. How can we determine the limit of (x2y)/(x4 + y2) as x and y approach infinity?

To determine the limit of (x2y)/(x4 + y2) as x and y approach infinity, we need to use algebraic manipulation and the properties of limits. By dividing both the numerator and denominator by the highest power of x, we can simplify the equation to (y)/(x2 + (y/x)2). As x approaches infinity, the term y/x will approach 0, making the denominator approach x2. Therefore, the limit as x and y approach infinity is 0.

3. Can we use the Squeeze Theorem to find the limit of (x2y)/(x4 + y2)?

Yes, we can use the Squeeze Theorem to find the limit of (x2y)/(x4 + y2). By setting x = y, we can rewrite the equation as (x2x)/(x4 + x2) = (x3)/(x2(x2 + 1)). As x approaches 0, the numerator and denominator both approach 0, making it suitable for the Squeeze Theorem. By using the fact that 0 ≤ x2 ≤ x2 + 1, we can squeeze the expression between 0 and x, resulting in a limit of 0.

4. How does the limit of (x2y)/(x4 + y2) change as x approaches 0?

As x approaches 0, the limit of (x2y)/(x4 + y2) remains 0. This can be seen by plugging in x = 0 into the equation and using algebraic manipulation and L'Hopital's rule to simplify it to (2y)/(4x3 + 2y). As x approaches 0, this expression will approach 0, resulting in a final limit of 0.

5. Can the limit of (x2y)/(x4 + y2) be evaluated using a graph or table?

No, the limit of (x2y)/(x4 + y2) cannot be evaluated using a graph or table. This is because the limit is a concept that involves the behavior of a function as the input values approach a certain value, rather than the actual output values of the function. A graph or table can only show the actual output values, not the behavior as the input values approach a certain value.

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