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Homework Help: Trouble with delta epsilon proofs

  1. Jan 1, 2010 #1
    i know how to do basic proofs, but some proofs on the actual limit theorems confuse me. my textbook's choices for delta are very obscure and i have no idea how they even came up with them.

    for the proof of the limit theorem where the limit of a product of 2 functions is equal to the product of their limits, my book did: f = L1 + (f-L1) and g = L2 + (g-L2). and they want to show that |f*g - L1*L2|< ε if 0<|x-a|<δ.

    so with substitution and rearrangement they get |L1(g-L2)+L2(f-L1)+(f-L1)(g-L2)|< ε. since the limit of f as x approaches a is L1 and limit of g as x approaches a is L2, we can find positive numbers δ1, δ2, δ3, δ4 such that:

    |f-L1|< sqrt(ε/3) if 0<|x-a|<δ1
    |f-L1|< ε/[3(1+|L2|)] if 0<|x-a|<δ2
    |g-L2|< sqrt(ε/3) if 0<|x-a|<δ3
    |g-L2|< ε/[3(1+|L1|)] if 0<|x-a|<δ4

    the remainder of the proof after the above step i understand but what confuses me is where and how did they get those expressions like sqrt(ε/3) and ε/[3(1+|L2|)]?
     
  2. jcsd
  3. Jan 1, 2010 #2
    you're not supposed to fix delta at the start. suppose you want to find a delta such that expression1<delta implies expression2<epsilon:

    start with:
    expression1 < delta

    then:
    -manipulate inequalities-
    -manipulate inequalities-
    -manipulate inequalities-


    and suppose you end up with, for example,
    expression2 < 3*delta^2


    Then you just look at the final inequality and go, 'if i make delta=sqrt(ε/3)' then expression2 < ε.
     
  4. Jan 1, 2010 #3

    Dick

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    |L1(g-L2)+L2(f-L1)+(f-L1)(g-L2)|<=|L1(g-L2)|+|L2(f-L1)|+|(f-L1)(g-L2)|. They conveniently choose those deltas so that each term in the sum is less than epsilon/3. So the total sum is less that epsilon. For example, the first term is less than |L1|*(epsilon/(3(1+|L1|)). |L1|/(1+|L1|)<=1. So the product is less than epsilon/3. The sqrt(epsilon/3) inequalities come in handy for the last term.
     
    Last edited: Jan 1, 2010
  5. Jan 1, 2010 #4
    how did the book get those 4 inequalities though? the expressions seem very random to me. sqrt(ε/3) and ε/[3(1+|L1|)] and ε/[3(1+|L2|)]. the book simply just wrote those 4 inequalities in the proof without showing how they got those expressions at all. that's what i'm confused about.
     
  6. Jan 1, 2010 #5

    Dick

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    They are random in the sense that they aren't the only choice you can make. Many choices will work. They are just one of them.
     
  7. Jan 1, 2010 #6
    ok I understand how the 3 terms reduce down to ε/3 and add up together to ε in the end. so did they just kind of guess and check to see which deltas would work out like that? I followed the proof step by step from beginning to end and I couldn't figure out how they just produced those inequalities seemingly out of nowhere.
     
  8. Jan 1, 2010 #7
    oh ok I understand now. thank you both!!
     
  9. Jan 1, 2010 #8

    Dick

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    Yes, they just looked at the first term and said what's a delta choice that will make that less than epsilon/3. Same for the other terms.
     
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