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Proving the limit of a multivariable function

  1. Jan 31, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the limit if it exists, or show that the limit does not exist.
    lim (x,y)-> (1,0) (xy-y)/((x-1)^2+y^2)

    2. Relevant equations
    lim (x,y)-> (a,b) f(x,y)
    0<((x-a)^2+(y-b)^2)^1/2<[itex]\delta[/itex]
    abs(f(x,y)-L)<[itex]\epsilon[/itex]

    3. The attempt at a solution
    I tried to prove that it does not exist by analyzing the limit coming in from the x & y axes, and along lines y=x, yatta yatta. I kept getting 0, so I then tried to prove the limit exists and equals zero using the delta epsilon method. There I ran into problems, I have a total of 3 calculus books, each only has one example for the method and they are all the same example, which is also the same and only example that was covered in my class. lim (x,y) -> (0,0) (3yx^2)/(x^2+y^2). I am just looking for a starting point.

    So far I have
    0<((x-1)^2+(y)^2)^1/2<[itex]\delta[/itex]
    abs((xy-y)/((x-1)^2+y^2))<[itex]\epsilon[/itex]
    I know I need to manipulate it so that I can relate delta to be some multiple of epsilon, but don't know how.
     
  2. jcsd
  3. Jan 31, 2013 #2

    SammyS

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    Try another path.

    y = m(x-1) approaches the point (1, 0) along a line of arbitrary slope.
     
  4. Jan 31, 2013 #3

    Dick

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    I'd first try a change of variables. Put u=x-1. So the limit is now (u,y)->(0,0). Denominator becomes u^2+y^2. What's the numerator? It might be easier to see that way.
     
    Last edited: Jan 31, 2013
  5. Jan 31, 2013 #4
    Thank both of you.
     
  6. Jan 31, 2013 #5
    Would either of you mind giving an example of using the delta epsilon method anyway?
     
  7. Jan 31, 2013 #6

    Dick

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    You don't need it for this one. It has no limit.
     
  8. Jan 31, 2013 #7
    Would either of you mind giving a different* example of using the delta epsilon method anyway?
     
  9. Feb 1, 2013 #8

    Dick

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    Ok, take f(x,y)=4xy/sqrt(x^2+y^2). (x,y)->(0,0). Change it to polar coordinates. You get |4r*cos(θ)*r*sin(θ)/r|=|4r*cos(θ)*sin(θ)|<=4r. So the limit is 0. And r is the "((x-a)^2+(y-b)^2)^1/2" in your definition. Pick ε>0. You want |f(x,y)-0|<ε, |f(x,y)-0|<4r, so if you pick δ=ε/4 it works.
     
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