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

[physics=world]

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?

 

[electricspit]

Try approaching the problem using polar coordinates:

$x=r\cos{\theta}$
$y=r\sin{\theta}$

 

[physics=world]

Try approaching the problem using polar coordinates:

$x=r\cos{\theta}$
$y=r\sin{\theta}$

After solving I get cos²(theta) / sin(theta)

 

[STEMucator]

You could choose $\delta = min\{1,2 \epsilon\}$

 

[electricspit]

Uh according to what I have, I'll do the numerator and let you do the denominator:

$x^2 y = r^3 \cos{\theta}^2\sin{\theta}$

Also remember $(x,y)\to 0$ is the same as $r\to 0$

 

[physics=world]

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?

 

[electricspit]

I get the limit to exist. Make sure you are making the substitutions correctly.

 

[physics=world]

Uh according to what I have, I'll do the numerator and let you do the denominator:

$x^2 y = r^3 \cos{\theta}^2\sin{\theta}$

Also remember $(x,y)\to 0$ is the same as $r\to 0$

denominator:

x⁴ + y² = r⁴cos⁴(theta) + sin²(theta)

 

[electricspit]

Ah see you are missing the r squared term in front of your sine term.

 

[SammyS]

I get the limit to exist. Make sure you are making the substitutions correctly.

I get that the limit doesn't exist.

 

[electricspit]

http://m.wolframalpha.com/input/?i=lim+x+to+0,+y+to+0+of+(x^2+y)/(x^4+y^2)&x=0&y=0

 

[SammyS]

Let (x, y) → (0, 0) along either of the paths, y = x² and y = -x² .

 

[electricspit]

Touché, WolframAlpha lies, as my polar coordinate method is not sufficient.

 

[Ray Vickson]

After solving I get cos²(theta) / sin(theta)

You shouldn't. You made a mistake.

 

[SammyS]

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.

 

[electricspit]

Yeah I just used the polar coordinates method and assumed Wolfram would have given the correct answer.