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Delta Epsilon Limit Proof - f(x,y) = xy

  1. Oct 19, 2008 #1
    How does one prove:

    limit xy = ab
    x-> a
    y -> b

    using the precise definition of a limit?

    My attempt:

    |xy-ab|<ϵ

    for:

    0<|x-a|<δ/2
    0<|y-b|<δ/2

    it follows that:

    δ/2-a <x< δ/2+a
    δ/2-b <y< δ/2+b

    then:

    (δ/2-a)(δ/2-n) < xy < (δ/2+a)(δ/2+b)

    (δ^2/4-aδ/2-bδ/2 +ab) <xy< (δ^2/4+aδ/2+bδ/2+ab )

    So -

    |xy-ab|< (δ^2/4+aδ/2+bδ/2)

    and I need to make

    |xy-ab|<ϵ

    At this point I feel a bit lost - anyone have any ideas for this proof?

    Thanks,

    Jeff
     
  2. jcsd
  3. Oct 20, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, you want [itex]|xy- ab|< \epsilon[/itex] . Now, xy- ab= xy- ay+ ay- ab= y(x-a)+ a(y-b) so [itex]|xy- ab|\le |y||x-a|+ |a||y- b|[/itex]. Since y converges to b, you can make |y-b| as small as you like so you can make it less than [itex]\epsilon/2a[/itex]. The other term is a bit tricky. You need to make |y||x-a| less than [itex]\epsilon/2[/itex] but y is a variable. However, since y converges to b, you can make |y| less than |b|+1 so you can use that instead: since x converges to a, you can make [itex]|x-a|\le \epsilon/(|b|+ 1)< \epsilon/|y|[/itex].
     
  4. Oct 20, 2008 #3
    Any idea what's happening with the latex output?
     
  5. Oct 21, 2008 #4
    ( I assume a,b ≠ 0 )

    | x - a | < ε' hence | x∙y - a∙y | < ε' ∙ | y |
    | y - b | < ε' hence | y∙a - b∙a | < ε' ∙ | a |

    you sum both inequalities

    | x∙y - a∙y | + | y∙a - b∙a | < ε' ∙ | y | + ε' ∙ | a |

    you use the triangle inequality

    | x∙y - a∙y | + | y∙a - b∙a | > | x∙y - a∙y + y∙a - b∙a | = | x∙y - a∙b |

    define ε = ε' ∙ | y | + ε' ∙ | b |

    now putting everything together yields : ...
     
    Last edited: Oct 21, 2008
  6. Oct 21, 2008 #5
    maybe multiplying works too?

    -d1 + a < x < d1 + a
    -d2 + b < y < d2 + b

    =>
    since -d1d2 < d1d2,

    -d1d2-d1b-d2a < d1d2-d1b-d2a < xy - ab < d1d2 + d1b + d2a

    then,

    abs(xy - ab) < d1d2 + d1b + d2a

    since a and b are fixed, as d1 and d2 go to zero, the abs goes to zero.

    [edit: hm. no, the sign on the multiplication makes this an error.]
     
    Last edited: Oct 21, 2008
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