Prove limit theorem using epsilon-delta

In summary, the conversation discusses Spivak's proof of the limit of a product of two functions, where one function has a limit of l and the other has a limit of m. The proof involves combining two delta values, \delta_1 and \delta_2, into one delta value, \delta, by taking the minimum of the two. However, there is confusion about why Spivak adds a 1 to \delta_2, and it is suggested that it may be to avoid dividing by 0. It is also mentioned that stating m\neq0 and l\neq0 could be an alternative to adding the 1.
  • #1
jessjolt2
16
0
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:

[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 want to 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 can't 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?
 
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  • #2
jessjolt2 said:
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:

[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 want to 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 can't 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
micromass said:
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 doesn't get [itex]\left|f(x)-l\right|[/itex] < min (1, E/( 2(|m|) )

He is adding some random 1 next to the |m| ?
 
  • #4
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
micromass said:
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
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
micromass said:
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[itex]\neq0[/itex], l[itex]\neq0[/itex] ?

i think the 1 comes from a more logical step?
 
  • #8
jessjolt2 said:
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
micromass said:
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 can't possibly follow from some earlier step?? i just want to make sure because it is bothering me lol
 
  • #10
jessjolt2 said:
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.
 
  • #11
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.
 

Related to Prove limit theorem using epsilon-delta

1. What is the epsilon-delta definition of a limit?

The epsilon-delta definition of a limit is a mathematical method used to formally define the concept of a limit in calculus. It states that for a given function f(x) and a limit L, for any positive value of epsilon (ε), there exists a positive value of delta (δ) such that if the distance between the input x and the limit L is less than delta (|x - L| < δ), then the difference between the output f(x) and the limit L is less than epsilon (|f(x) - L| < ε).

2. How do you prove a limit using epsilon-delta?

To prove a limit using epsilon-delta, you must first choose a value of epsilon (ε) and then find a corresponding value of delta (δ) that satisfies the definition of a limit. This is done by manipulating the function and the limit algebraically until the desired inequality is achieved. It is important to note that the value of delta may vary depending on the chosen value of epsilon. Once you have found a suitable delta, you must then prove that it works for all x values within the given interval.

3. What is the importance of the epsilon-delta definition of a limit?

The epsilon-delta definition of a limit is important because it provides a rigorous and precise way to define the concept of a limit. It allows us to prove the existence of a limit and determine its value with complete certainty, rather than relying on intuition or graphical approximations. This definition is also essential in the development of advanced calculus concepts such as continuity and differentiability.

4. What is the role of epsilon and delta in the definition of a limit?

Epsilon (ε) and delta (δ) are both variables used in the epsilon-delta definition of a limit. Epsilon represents the desired level of precision or closeness to the limit, while delta represents the corresponding range of values for the input that will produce an output within that desired level of precision. In other words, epsilon and delta determine the conditions that must be met for a limit to exist.

5. Can the epsilon-delta definition of a limit be applied to all functions?

Yes, the epsilon-delta definition of a limit can be applied to all functions as long as the limit and the function are well-defined and continuous. However, some functions may be more difficult to prove using this method, and alternative methods such as the squeeze theorem may be used instead. It is also important to note that the values of delta and epsilon may vary depending on the function and the chosen limit.

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