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Proof of a limit

  1. Jun 15, 2010 #1
    Delta epsilon definition of a limit and its proof

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

    Prove; lim x2+y2= a2+b2
    (x,y) ---->(a,b)


    2. Relevant equations

    Not much equations are present, just some algebra, and the definition of a limit.


    3. The attempt at a solution

    Well, this is what I did

    (x2+y2)-(a2+b2) < \sqrt{(x-a)2+(y-b)2} = or < than \delta.

    so i opened the right hand, and left hand of the equation with the square root out.

    x2+y2-a2-b2=(x2+y2+a2+b2-2xa-2yb)-2a2-2b2+2xa+2yb

    The parenthesis in the right hand side is /delta2. and all the other stuff on the right hand side, I'm going to try to write it in terms of \delta .

    -2a2-2b2+2xa+2yb= (x-a)+(y-b)

    so, I fix these up a bit;

    (x-a)+(y-b)=2a(x-a) + 2b(y-b) < \delta

    so here, 2a(x-a) is 2a\delta and 2b(y-b) is 2b\delta

    and we know that x-a<\delta and y-a<\delta

    so in the worst case, lets equal these pieced to \delta, we will have

    2a\delta + 2b\delta + \delta2 <\epsilon


    And from here, I just solve for \delta.

    Is this the correct way to solve this? Lend me a helping hand the masters of math.
     
    Last edited: Jun 15, 2010
  2. jcsd
  3. Jun 17, 2010 #2

    lanedance

    User Avatar
    Homework Helper

    Re: Delta epsilon definition of a limit and its proof

    this statement isn't true
     
  4. Jun 17, 2010 #3
    well that first statement is from the epsilon-delta definition of a limit. I know they don't equal each other, but they do have a relationship, and that's how you approach the problem, that's what I think.
     
  5. Jun 17, 2010 #4
    Re: Delta epsilon definition of a limit and its proof

    To better understand the definition of a multiviariable func

    here is a step-by-step guide http://www.math.utah.edu/~arcara/teaching/0308/handout.pdf
     
  6. Jun 17, 2010 #5
    Suzanne, the link that you posted doesn't say how to prove that a limit exists, it basically says that we did such things and now do these for practice.

    Is there no one here that knows about these things in detail?
     
  7. Jun 17, 2010 #6

    Mark44

    Staff: Mentor

    You need to show that for any positive epsilon, there is a positive number delta such that if |(x, y) - (a, b)| < delta, then |x^2 + y^2 - a^2 - b^2)| < epsilon.

    What the first inequality says is that (x, y) can be any point within delta of the point (a, b). I'm having a hard time following your work, especially the part just before you said you "opened" the right side.
     
  8. Jun 17, 2010 #7
    well the right hand side is the length. [(x-a)^2+(y-b)^2]^.5, When I say I open this up, I mean, i take its square and distribute the squares.
     
  9. Jun 19, 2010 #8

    Gib Z

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    Homework Helper

    It's extremely hard to read your post since you don't use latex, so I'm going to try lead you down a different way.

    You want to show that for every ε>0 there exists δ>0 so that [tex](x-a)^2 + (y-b)^2 < \delta^2[/tex] implies [tex] ( x^2-a^2 + y^2-b^2) = (x-a)(x+a) + (y-b)(y+b) < \epsilon [/tex].

    So a method you could use is to start with [tex](x-a)^2 + (y-b)^2 < \delta^2[/tex], and use it to show that it implies some other inequality [tex](x-a)(x+a) + (y-b)(y+b) < f(\delta)[/tex]. Then if you can show that f is invertible, so that for every ε>0 there exists δ>0 so that f(δ) = ε, then we have found a way to satisfy our definition. The first step in this method is to use the starting inequality and get some basic, weaker inequalities from it, so do that first and play around with it.
     
  10. Jun 19, 2010 #9

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    With functions of two variables, I find it easiest to "translate" so that the limit is as (x, y) goes to (0, 0) and then change to polar coordinates. That way the distance to (0, 0) is measured by the single variable, r.

    Let x'= x- a, y'= y- b. Then x= x'+ a, y= y'+ b, and [itex]x^2+ y^2= (x'+a)^2+ (y'+ b)^2=[/itex][itex] x'^2+ y'^2+ 2ax'+ 2by'+ a^2+ b^2[/itex]. In polar coordinates, [itex]x'^2+ y'^2= r^2[/itex] [itex]2ax'= 2ar cos(\theta)[/itex] and [itex]2by'= 2br sin(\theta)[/itex] so that [itex]x^2+ y^2= r^2+ 2r(a cos(\theta)+ b sin(\theta)+ a^2+ b^2[/itex].

    That will have limit, as (x, y) goes to (a, b), [itex]a^2+ b^2[/itex] if and only if [itex]r^2+ 2r(a cos(\theta)+ b sin(\theta)= r(r+ 2(acos(\theta)+ b sin(\theta))[/itex] goes to 0 as r goes to 0 (for any value of [itex]\theta[/itex]).
     
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