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Multivariable Calculus Limit process

  • #1

Homework Statement



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

Homework 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)

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.
Thank you for your help.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi theno1katzman! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
take the limit as (x,y) -> (0,0) of f(x,y) where f(x,y) = (x^6-y^6)/(x^3-y^3)

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?
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. :wink:
 
  • #3
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
tiny-tim
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Homework Helper
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Thanks Tim, when I tried approaching (0,0) along the curve y=e^x …
(please use the X2 tag just above the Reply box :wink:)

erm :redface: … noooo …

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

Try again, to find a curve on which f(x,y) is constant (you may have to solve a quadratic). :smile:
 
Last edited:
  • #5
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
tiny-tim
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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
[tex]y = x^6 + c [/tex] [tex] lim (x,y) -> (0,0) f(x,y) [/tex]
 
  • #8
24
0
Did you notice that the numerator is a difference of squares ?
 
  • #9
Ahh yes I will simplify this to [tex]x^3+y^3[/tex] and work the problem again
 
  • #10
24
0
Ahh yes I will simplify this to [tex]x^3+y^3[/tex] and work the problem again
So the limit = ?

The polar coordinates works also :)
 
  • #11
limit would be 0 if I use polar coordinates, question, would polar coordinates prove the limit exists for all paths?
 
  • #12
24
0
limit would be 0 if I use polar coordinates, question, would polar coordinates prove the limit exists for all paths?
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
Thank you, I wish my professor would have clarified that in class.
 

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