Trouble with delta epsilon proofs

  • #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|)]?
 

Answers and Replies

  • #2
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0
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.
 
  • #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:
  • #4
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.

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.
 
  • #5
Dick
Science Advisor
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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.

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.
 
  • #6
|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 that 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.

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.
 
  • #7
oh ok I understand now. thank you both!!
 
  • #8
Dick
Science Advisor
Homework Helper
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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.

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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