Prove limit theorem using epsilon-delta

[jessjolt2]

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 don't understand why he is doing something..

ok so the following i understand:

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

\left|f(x)-l\right|\leq1
\left|f(x)\right|\leq\left|l\right|+1

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

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

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


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

cus i mean m is just a constant..it can't be restricted like f(x) was...

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

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

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



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

 

[micromass]

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 don't understand why he is doing something..

ok so the following i understand:

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

\left|f(x)-l\right|\leq1
\left|f(x)\right|\leq\left|l\right|+1

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

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

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


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

cus i mean m is just a constant..it can't be restricted like f(x) was...

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

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

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

Yes, but now you have done something like

|f(x)-l|&lt;\delta_1~\vert~|g(x)-m|&lt;\delta_2

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

So Spivak has two delta's:

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

and he combines it to one delta. This is what he does.

 

[jessjolt2]

Yes, but now you have done something like

|f(x)-l|&lt;\delta_1~\vert~|g(x)-m|&lt;\delta_2

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

So Spivak has two delta's:

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

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 \left|f(x)-l\right| < min (1, E/( 2(|m|+1) ) ) ?

He doesn't get \left|f(x)-l\right| < min (1, E/( 2(|m|) )

He is adding some random 1 next to the |m| ?

 

[micromass]

He's combining the delta's

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

into

\delta=\min(\delta_1,\delta_2)=\min(1,\frac{1}{2|m|+1})

 

[jessjolt2]

He's combining the delta's

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

into

\delta=\min(\delta_1,\delta_2)=\min(1,\frac{1}{2|m|+1})

But delta#2 is not \frac{1}{2|m|+1}

it is \frac{1}{2(|l|+1)}

 

[micromass]

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.

 

[jessjolt2]

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 don't think u can just add a random 1 without explaining why?

why not just say that m\neq0, l\neq0 ?

i think the 1 comes from a more logical step?

 

[micromass]

haha its legal to just do that?

It is in this case, since

\frac{1}{2(|l|+1)}&lt;\frac{1}{2|l|}

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

why not just say that m\neq0, l\neq0 ?

This is also ok, but then you would have to give a different proof in the case that l=0 or m=0.

 

[jessjolt2]

It is in this case, since

\frac{1}{2(|l|+1)}&lt;\frac{1}{2|l|}

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 can't possibly follow from some earlier step?? i just want to make sure because it is bothering me lol

 

[micromass]

ohhh gotcha thanks...are you sure though that this 1 can't 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.

 

[deluks917]

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+\delta)(b+\delta) - ab | < \epsilon

| (a+b)\delta + \delta²| < \epsilon

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

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