Prove limit theorem using epsilon-delta

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Main Question or Discussion Point

Hey i am trying to understand Spivak's proof of lim x->a of f(x)g(x)=lm (where l is limit of f(x) and m is lim of g(x) )..but i think he is skipping many steps and at one point i dont understand why he is doing something..

ok so the following i understand:

[itex]\left|f(x)g(x)-lm\right|[/itex]< E
[itex]\leq[/itex][itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex] + [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E

[itex]\left|f(x)-l\right|[/itex][itex]\leq[/itex]1
[itex]\left|f(x)\right|[/itex][itex]\leq[/itex][itex]\left|l\right|[/itex]+1

[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex]<([itex]\left|l\right|[/itex]+1)[itex]\left|g(x)-m\right|[/itex]<E/2
[itex]\left|g(x)-m\right|[/itex]< [itex]\frac{E}{2(\left|l\right|+1)}[/itex]

[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex]<([itex]\left|l\right|[/itex]+1)[itex]\frac{E}{2(\left|l\right|+1)}[/itex]

[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex]<E/2


ok so now i am guessing that we wanna say that [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E/2 so that E/2 + E/2 = E
so can u just say:
[itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E/2
[itex]\left|f(x)-l\right|[/itex]<E/(2[itex]\left|m\right|[/itex])

cus i mean m is just a constant..it cant be restricted like f(x) was...

so:
[itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<[itex]\left|m\right|[/itex](E/(2[itex]\left|m\right|[/itex]))= E/2

so then:
[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex] + [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]< E/2 + E/2 =E

when [itex]\left|f(x)-l\right|[/itex] < min (1, E/(2|m|) ) and [itex]\left|g(x)-m\right|[/itex]<[itex]\frac{E}{2(\left|l\right|+1)}[/itex]



but Spivak is saying that [itex]\left|f(x)-l\right|[/itex] < min (1, E/( 2(|m|+1) ) ) and i have no clue why... help plsss?
 

Answers and Replies

  • #2
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Hey i am trying to understand Spivak's proof of lim x->a of f(x)g(x)=lm (where l is limit of f(x) and m is lim of g(x) )..but i think he is skipping many steps and at one point i dont understand why he is doing something..

ok so the following i understand:

[itex]\left|f(x)g(x)-lm\right|[/itex]< E
[itex]\leq[/itex][itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex] + [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E

[itex]\left|f(x)-l\right|[/itex][itex]\leq[/itex]1
[itex]\left|f(x)\right|[/itex][itex]\leq[/itex][itex]\left|l\right|[/itex]+1

[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex]<([itex]\left|l\right|[/itex]+1)[itex]\left|g(x)-m\right|[/itex]<E/2
[itex]\left|g(x)-m\right|[/itex]< [itex]\frac{E}{2(\left|l\right|+1)}[/itex]

[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex]<([itex]\left|l\right|[/itex]+1)[itex]\frac{E}{2(\left|l\right|+1)}[/itex]

[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex]<E/2


ok so now i am guessing that we wanna say that [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E/2 so that E/2 + E/2 = E
so can u just say:
[itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<E/2
[itex]\left|f(x)-l\right|[/itex]<E/(2[itex]\left|m\right|[/itex])

cus i mean m is just a constant..it cant be restricted like f(x) was...

so:
[itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]<[itex]\left|m\right|[/itex](E/(2[itex]\left|m\right|[/itex]))= E/2

so then:
[itex]\left|f(x)\right|[/itex][itex]\left|g(x)-m\right|[/itex] + [itex]\left|m\right|[/itex][itex]\left|f(x)-l\right|[/itex]< E/2 + E/2 =E

when [itex]\left|f(x)-l\right|[/itex] < min (1, E/(2|m|) ) and [itex]\left|g(x)-m\right|[/itex]<[itex]\frac{E}{2(\left|l\right|+1)}[/itex]
Yes, but now you have done something like

[tex]|f(x)-l|<\delta_1~\vert~|g(x)-m|<\delta_2[/tex]

so you have two delta's. But you only want one delta. So if we take [itex]\delta=\min(\delta_1,\delta_2)[/itex], then [itex]|f(x)-l|<\delta[/itex] implies that [itex]|f(x)-l|<\delta_1[/itex]. And the same with g(x).

So Spivak has two delta's:

[tex]\delta=1=\min(1,\varepsilon/(2|m|))~\text{and}~\delta_2=1/(2(|l|+1))[/tex]

and he combines it to one delta. This is what he does.
 
  • #3
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Yes, but now you have done something like

[tex]|f(x)-l|<\delta_1~\vert~|g(x)-m|<\delta_2[/tex]

so you have two delta's. But you only want one delta. So if we take [itex]\delta=\min(\delta_1,\delta_2)[/itex], then [itex]|f(x)-l|<\delta[/itex] implies that [itex]|f(x)-l|<\delta_1[/itex]. And the same with g(x).

So Spivak has two delta's:

[tex]\delta=1=\min(1,\varepsilon/(2|m|))~\text{and}~\delta_2=1/(2(|l|+1))[/tex]

and he combines it to one delta. This is what he does.
I know that you have to combine the two deltas by taking the minimum delta, but my question is, how does Spivak get that [itex]\left|f(x)-l\right|[/itex] < min (1, E/( 2(|m|+1) ) ) ?

He doesnt get [itex]\left|f(x)-l\right|[/itex] < min (1, E/( 2(|m|) )

He is adding some random 1 next to the |m| ?
 
  • #4
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He's combining the delta's

[tex]\delta_1=\min(1,\frac{1}{2|m|})~\text{and}~\delta_2=\frac{1}{2|m|+1}[/tex]

into

[tex]\delta=\min(\delta_1,\delta_2)=\min(1,\frac{1}{2|m|+1})[/tex]
 
  • #5
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He's combining the delta's

[tex]\delta_1=\min(1,\frac{1}{2|m|})~\text{and}~\delta_2=\frac{1}{2|m|+1}[/tex]

into

[tex]\delta=\min(\delta_1,\delta_2)=\min(1,\frac{1}{2|m|+1})[/tex]
But delta#2 is not [itex]\frac{1}{2|m|+1}[/itex]

it is [itex]\frac{1}{2(|l|+1)}[/itex]
 
  • #6
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Ah yes, it seems you're right then. I suspect Spivak adds the +1 to make sure that he didn't divide by 0. That is: if l=0, then you would have 1/0 if you didn't add +1.
 
  • #7
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Ah yes, it seems you're right then. I suspect Spivak adds the +1 to make sure that he didn't divide by 0. That is: if l=0, then you would have 1/0 if you didn't add +1.
haha its legal to just do that? i mean this is supposed to be a rigid proof...i dont think u can just add a random 1 without explaining why?

why not just say that m[itex]\neq0[/itex], l[itex]\neq0[/itex] ???

i think the 1 comes from a more logical step?
 
  • #8
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haha its legal to just do that?
It is in this case, since

[tex]\frac{1}{2(|l|+1)}<\frac{1}{2|l|}[/tex]

So you're only making the delta smaller. This is certainly allowed.

why not just say that m[itex]\neq0[/itex], l[itex]\neq0[/itex] ???
This is also ok, but then you would have to give a different proof in the case that l=0 or m=0.
 
  • #9
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It is in this case, since

[tex]\frac{1}{2(|l|+1)}<\frac{1}{2|l|}[/tex]

So you're only making the delta smaller. This is certainly allowed.



This is also ok, but then you would have to give a different proof in the case that l=0 or m=0.
ohhh gotcha thanks....are you sure though that this 1 cant possibly follow from some earlier step?? i just want to make sure because it is bothering me lol
 
  • #10
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3,283
ohhh gotcha thanks....are you sure though that this 1 cant possibly follow from some earlier step?? i just want to make sure because it is bothering me lol
I looked at Spivak, and this is the only possible reason I see why he would want to add +1.
 
  • #11
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4
You should try to see why Spivak's proof works but when I learned this from Spivak it all felt very mysterious. See if you can do the proof yourself:

| (a+[itex]\delta[/itex])(b+[itex]\delta[/itex]) - ab | < [itex]\epsilon[/itex]

| (a+b)[itex]\delta[/itex] + [itex]\delta[/itex]2| < [itex]\epsilon[/itex]

can you see why its not too hard to find a [itex]\delta[/itex] that works.

----

I'm sorry if this post doesn't help but I never understood this proof unless I could find it out myself.
 

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