# Multivariable Calculus Limit process

1. Mar 5, 2010

### theno1katzman

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

Determine whether the limit exists; if it does, what is it?

2. Relevant equations

take the limit as (x,y) -> (0,0) of f(x,y) where f(x,y) = (x^6-y^6)/(x^3-y^3)

3. The attempt at a solution

What i started doing was approaching along the line y=0 and that would give
lim as (x,0) -> (0,0) x^6/x^3 simplifying to the lim as (x,0) -> (0,0) x^3 which is 0.

Then I approached along the line x=0 and that would give
lim as (0,y) -> (0,0) -y^6/-y^3 simplifying to the lim as (0,y) -> (0,0) y^3 which is 0.

Then i approached along the line y=x
lim as (x,x) -> (0,0) (x^6-x^6)/(x^3-x^3) = 0/0 which is an indeterminate form so I use L'Hospital's rule a few times and continue to get 0/0. This rational function is the same as saying (x-x)/(x-x).

I conclude that the limit does not exist becuase it can not be evaluated any further. Am i correct? Is there something else I shoudl be doing? This homework problem is due March 22.

2. Mar 5, 2010

### tiny-tim

Welcome to PF!

Hi theno1katzman! Welcome to PF!

(try using the X2 tag just above the Reply box )
hmm … I've a nasty feeling that the domain of the function isn't supposed to include the line x = y, in which case you can't approach along that line.

Try approaching along a curve on which f(x,y) is constant.

3. Mar 5, 2010

### theno1katzman

Thanks Tim, when I tried approaching (0,0) along the curve y=e^x I will get
limit of (x,e^x)->(0,0) of (x^6-e^(6 x))/(x^3-e^(3 x)) = 1

This is a different answer for the limit above which was 0, therefore the limit of the original function as (x,y) approaches (0,0) does not exist.

4. Mar 6, 2010

### tiny-tim

erm … noooo …

y = ex doesn't get anywhere near (0,0), does it?

Try again, to find a curve on which f(x,y) is constant (you may have to solve a quadratic).

Last edited: Mar 6, 2010
5. Mar 6, 2010

### theno1katzman

Let me know if this is correct. If I approached (0,0) along the path y=x n where n is any positive integer for example y=x2 then after factoring I would get

lim (x,x2)-> (0,0) x3(x2-x+1)/(x2+x+1) = 0/1 = 0.

Therefore the limit exists and is 0.

6. Mar 6, 2010

### tiny-tim

I didn't follow that … for y = x2, it would be x3(1 - x6)/(1 - x3), which does --> 0.

But that doesn't prove it for all paths, does it?

Try my previous hint, to find a curve on which f(x,y) is constant.

7. Mar 6, 2010

### theno1katzman

$$y = x^6 + c$$ $$lim (x,y) -> (0,0) f(x,y)$$

8. Mar 6, 2010

### Sweet_GirL

Did you notice that the numerator is a difference of squares ?

9. Mar 6, 2010

### theno1katzman

Ahh yes I will simplify this to $$x^3+y^3$$ and work the problem again

10. Mar 6, 2010

### Sweet_GirL

So the limit = ?

The polar coordinates works also :)

11. Mar 6, 2010

### theno1katzman

limit would be 0 if I use polar coordinates, question, would polar coordinates prove the limit exists for all paths?

12. Mar 6, 2010

### Sweet_GirL

Yes.
There is a big difference in solving multivariable limits between the 2-path rule (which is y=mx .. etc) and the polar coordinates method.
2-path rule proves only that the limit D.N.E and does not prove the existence of the limit.
Polar coordinates method proves both cases.

13. Mar 6, 2010

### theno1katzman

Thank you, I wish my professor would have clarified that in class.