# Find the limit

1. Apr 9, 2014

### physics=world

1. The problem statement, all variables and given/known data

Find the limit:

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

2. Relevant equations

3. 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?

2. Apr 9, 2014

### electricspit

Try approaching the problem using polar coordinates:

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

3. Apr 9, 2014

### physics=world

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

4. Apr 9, 2014

### Zondrina

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

5. Apr 9, 2014

### 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 $r\to 0$

6. Apr 9, 2014

### 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?

7. Apr 9, 2014

### electricspit

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

8. Apr 9, 2014

### physics=world

denominator:

x4 + y2 = r4cos4(theta) + sin2(theta)

9. Apr 9, 2014

### electricspit

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

10. Apr 9, 2014

### SammyS

Staff Emeritus
I get that the limit doesn't exist.

11. Apr 9, 2014

### electricspit

12. Apr 9, 2014

### SammyS

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

13. Apr 9, 2014

### electricspit

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

14. Apr 9, 2014

### Ray Vickson

You shouldn't. You made a mistake.

15. Apr 9, 2014

### SammyS

Staff Emeritus
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. Apr 10, 2014

### electricspit

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