Limits Question

  • Thread starter Paradox
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  • #1
How can one prove that:

lim (x,y)->(0,0)

(x^4+y^4)
---------
(x^2+y^2)

= 0

I keep getting 0/0 no matter what I do to the equation. Anyone have any pointers?
 
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Answers and Replies

  • #2
Hurkyl
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Apply the epsilon-delta definition of a limit, or try a clever substitution.

Hurkyl
 
  • #3
Thank you for the quick response. I'll go and try it out now.
 
  • #4
HallsofIvy
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One of the difficulties with limits in more than one variable is that there are an infinite number of different ways to "approach" the target point. In order for the limit to exist, the result must be the same along any path.


Since the "epsilon-delta" definition Hurkyl mentioned used the distance from the target point the best way to use it is to change to polar coordinates so that r measures the distance from from (0,0).

Note that x= r cos([theta]) and y= r sin([theta]) so that
x<sup>2</sup>+ y<sup>2</sup>= r<sup>2</sup> and
x<sup>4</sup>+ y<sup>4</sup>= r<sup>4</sup>(cos<sup>4</sup>([theta])+sin<sup>4</sup>([theta]).

That should make it easy.
 
  • #5
HallsofIvy
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Oops, wrong forum, wrong symbols. Well, the math is still correct.
 
  • #6
Loren Booda
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Would using L'Hospital's rule (successive differentiations of numerator and denominator) be cheating?
 
  • #7
HallsofIvy
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Since this is a function of two variables, HOW, exactly, would you apply L'Hospital's rule?
 
  • #8
Thanks HallsOfIvy. That was much easier to work with
 
  • #9
Loren Booda
3,119
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HallsofIvy
Since this is a function of two variables, HOW, exactly, would you apply L'Hospital's rule?
(d2N/dxNdyN)((x^4+y^4)/(x^2+y^2)),

where the derivatives are partial.
 
  • #10
ObsessiveMathsFreak
406
8
the proof is as follows.

lim(x,y)->(0,0)

x^4 + y^4
---------
x^2 + y^2

=

(x^2 + i*y^2)*(x^2 - i*y^2)
---------------------------
x^2 + y^2

=

x^2 - i*y^2

as x and y go to zero this value approches 0.

Not that when they are zero the value of the function is NOT zero.
 
  • #11
Loren Booda
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ObsessiveMathsFreak
Not that when they are zero the value of the function is NOT zero.
Is that value then undefined?
 
  • #12
bogdan
191
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Why

(x^2 + i*y^2)*(x^2 - i*y^2)
---------------------------
x^2 + y^2

=

x^2 - i*y^2

?

x^2 + i*y^2<>x^2 + y^2

Try this way :

(x^4+y^4)/(x^2+y^2)=((x^2+y^2)^2-2*x^2*y^2)/(x^2+y^2)=
=1-2*x^2*y^2/(x^2+y^2);
Now lim (x^2+y^2)-2*x^2*y^2/(x^2+y^2) = lim (x^2+y^2) -
lim 2*x^2*y^2/(x^2+y^2)= 0 -1 / lim (x^2+y^2)/2*x^2*y^2=
=-1 / lim (1/(2*x^2)+1/(2*y^2))=-1/infinity = 0;

I hope this is correct...

By the way...HallsofIvy...
I don't think your "notation" is correct...
Because if you say x=r*cost and y=r*sint you practically say
x=k*y, which is not correct...x could be equal to y^2...
(because x->0 and y->0 means that r->0, because cost and sint
can't -> 0 in the same time)
See ya...
 
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  • #13
HallsofIvy
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By the way...HallsofIvy...
I don't think your "notation" is correct...
Because if you say x=r*cost and y=r*sint you practically say
x=k*y, which is not correct...x could be equal to y^2...
(because x->0 and y->0 means that r->0, because cost and sint
can't -> 0 in the same time)

I SAID that was converting to polar coordinates. The point with coordinates r and theta in polar coordinates has x= r cos(t) and
y= r sin(t) in cartesian coordinates. Believe it or not I am completely aware that as (x,y)-> (0,0), r-> 0! That was the whole point! Since r measures the distance from (0,0) to the point, the two variables x and y going to 0 reduces to the single variable r going to 0.
 
  • #14
bogdan
191
0
(x,y) is a point ?
Not a pair of variables ?
My mistake then...
Sorry...
But there's no real need to consider them coord of a point...
 
  • #15
HallsofIvy
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If you are willing to ASSUME that you can you can
treat 1/(1/x^2+ 1/y^2) as x and y going to 0 as
"1/(infinity+infinity)" and declare that "1/infinity" is 0, then there is no real need to be precise at all. Yes, your method does work (and is very clever) in this example but I suspect that most mathematics professors would want you to show a little more understanding of what you are doing.
 

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