Proving the Limit of f(x)=0 in Calculus with Spivak's Problem

In summary: S) and a, we can choose a delta that is smaller than the minimum distance between those elements. And since S is finite, the minimum distance is positive, so we can choose a delta that is positive. Therefore, the limit as x approaches a of f is equal to 0.
  • #1
Bleys
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I came across a problem in Calculus by Spivak, and I'm having trouble formalizing the proof.
Let [tex]A_{n}[/tex] bet a set of finite numbers in [0,1], and if [tex]m \neq n[/tex] then [tex]A_{n}[/tex] and [tex]A_{m}[/tex] are disjoint. Let f(x) be defined as
f(x)=1/n if x is in [tex]A_{n}[/tex] and f(x)=0 if x is not in any [tex]A_{n}[/tex]. The question asks to prove the limit as x goes to a of f is 0 for any a in [0,1].
Now I thought: given an n, there are only finitely many elements of [tex]A_{n}[/tex] in a neighborhood of a. Choose the smallest such n, say [tex]n_{0}[/tex]. Then [tex]f(x)\leq1/n_{0}[/tex]. Restrict the neighborhood further so that none of the elements of [tex]A_{n_{0}}[/tex] are in the interval. Then choose the next n such that it's minimal. Obviously [tex]f(x)\leq1/n\leq1/n_{0}[/tex]. Successively doing this, for arbitrarily small x, f(x) will tend to 0.
I don't know how to prove this using the limit definition; can someone help me out with what [tex]\delta[/tex] to choose?
 
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  • #2
The smallest such n with what property? You're not very clear about that. Here's a proof: I'm going to assume f(a)=0; the other case is only a slight modification and should be instructive for you to think about. Fix [tex]\varepsilon>0[/tex], then there exists N > 0 with [tex]1/N < \varepsilon[/tex]. Let [tex]B_N = \cup_{ n = 1 }^N A_n[/tex], so B_N is a finite set, and [tex]a \not \in B_N[/tex] (why?). Now, let [tex]\delta > 0[/tex] be such that [tex]( a - \delta, a + \delta ) \cap B_N = \emptyset[/tex] (why does such a [tex]\delta[/tex] exist?). Then if [tex]| a - x | < \delta[/tex], either [tex]x \in A_m[/tex] for m > N, or f(x) = 0. In either case, [tex]|f(a) - f(x)|<1/N<\varepsilon[/tex], as desired.
 
  • #3
Bleys said:
I came across a problem in Calculus by Spivak, and I'm having trouble formalizing the proof.
Let [tex]A_{n}[/tex] bet a set of finite numbers in [0,1], and if [tex]m \neq n[/tex] then [tex]A_{n}[/tex] and [tex]A_{m}[/tex] are disjoint. Let f(x) be defined as
f(x)=1/n if x is in [tex]A_{n}[/tex] and f(x)=0 if x is not in any [tex]A_{n}[/tex]. The question asks to prove the limit as x goes to a of f is 0 for any a in [0,1].
Now I thought: given an n, there are only finitely many elements of [tex]A_{n}[/tex] in a neighborhood of a. Choose the smallest such n, say [tex]n_{0}[/tex]. Then [tex]f(x)\leq1/n_{0}[/tex]. Restrict the neighborhood further so that none of the elements of [tex]A_{n_{0}}[/tex] are in the interval. Then choose the next n such that it's minimal. Obviously [tex]f(x)\leq1/n\leq1/n_{0}[/tex]. Successively doing this, for arbitrarily small x, f(x) will tend to 0.
I don't know how to prove this using the limit definition; can someone help me out with what [tex]\delta[/tex] to choose?

you're on the right track, you should break this down into 2 parts:
1) if a is in a set An
2) if a is not in a set An

given an epsilon larger than 0, you must consider all f(x) = 1/n that fails ( larger than epsilon). Suppose you consider all 1/n/f(x) >= epsilon, then you can consider a union of sets (S) An such that f(x) >= epsilon for any element in a set An. Now you've reduced the problem to taking delta to be the min distance between the members of that set (S) and a (the distance between the greatest lower bound of (S) and a). If a is a member of the set itself, then you can use the exact same argument, since we are only concerned with behaviour as x APPROACHES a. Also, you know that you can always use this general argument because the interval between any numbers is always infinitely dense.
 
  • #4
The smallest such n with what property? You're not very clear about that
Sorry, what I meant to say was the smallest n such that the set [tex]A_{n}[/tex] has an element in the neighborhood of a.
so B_N is a finite set, and (why?)
f(a)=0 iff a=0, by definition of f, and this is iff a is not in any An.
(why does such a [tex] \delta [/tex] exist?)
Because B_N is finite.

Did you assume f(a)=0 because you had [tex]( a - \delta, a + \delta )[/tex]? Would [tex]0 < | a - x | < \delta[/tex] solve that? After all, since we are looking at the limit 0 then the expression [tex]|f(a) - f(x)|<\varepsilon[/tex] becomes [tex]|f(x) - 0|=|f(x)|<\varepsilon[/tex], since the value of f at a doesn't matter.

given an epsilon larger than 0, you must consider all f(x) = 1/n that fails ( larger than epsilon). Suppose you consider all 1/n/f(x) >= epsilon, then you can consider a union of sets (S) An such that f(x) >= epsilon for any element in a set An. Now you've reduced the problem to taking delta to be the min distance between the members of that set (S) and a
So since S will be finite (since there will be an n such that [tex]1/n<\varepsilon[/tex]) then that delta will work. Can I just not consider whether a is in the set or not. After all like you said, we're interested in the limit, not continuity. So [tex]0 < | a - x | < \delta[/tex] would solve that.
Ok I think I get it. I was having trouble formulating all this with the [tex]\delta - \varepsilon[/tex] definition. Thanks for all your help!
 
  • #5
Bleys said:
Sorry, what I meant to say was the smallest n such that the set [tex]A_{n}[/tex] has an element in the neighborhood of a.
f(a)=0 iff a=0, by definition of f, and this is iff a is not in any An.
Because B_N is finite.

Did you assume f(a)=0 because you had [tex]( a - \delta, a + \delta )[/tex]? Would [tex]0 < | a - x | < \delta[/tex] solve that? After all, since we are looking at the limit 0 then the expression [tex]|f(a) - f(x)|<\varepsilon[/tex] becomes [tex]|f(x) - 0|=|f(x)|<\varepsilon[/tex], since the value of f at a doesn't matter.So since S will be finite (since there will be an n such that [tex]1/n<\varepsilon[/tex]) then that delta will work. Can I just not consider whether a is in the set or not. After all like you said, we're interested in the limit, not continuity. So [tex]0 < | a - x | < \delta[/tex] would solve that.
Ok I think I get it. I was having trouble formulating all this with the [tex]\delta - \varepsilon[/tex] definition. Thanks for all your help!

It's a good idea to say how the limit works in both cases, you need to show that you've considered that case (it is a significant case). But in the end, it uses the same argument.
The delta you are taking is the min ( |a-x| : x [tex]\in[/tex] [tex]\cup[/tex] n=1 to i An )

Here is an intuitive version of the proof:

|A1||A15||A20||A4|<----a------>|A7||A9||A10000||A124921894381241|

Any x's will be dispersed into different sets An, so take delta to be the min distance between a set and the point a. This interval between a and any x in sets An will always exist, since an interval of real numbers is always infinitely dense
 

Related to Proving the Limit of f(x)=0 in Calculus with Spivak's Problem

What is Spivak's Problem in Calculus?

Spivak's Problem is a mathematical exercise in calculus that asks students to prove that the limit of a function f(x) is equal to 0. It is often used as a challenging problem for students to practice their understanding of the concept of limits.

How do you prove the limit of f(x) is 0 using Spivak's Problem?

To prove the limit of f(x) is 0 using Spivak's Problem, you must show that for any given positive number ε, there exists a corresponding number δ such that when the distance between x and a (the point at which the limit is being evaluated) is smaller than δ, the distance between f(x) and 0 is smaller than ε. This can be done through a rigorous mathematical proof.

Why is Spivak's Problem considered difficult?

Spivak's Problem is considered difficult because it requires a deep understanding of the concept of limits and the ability to use mathematical logic to construct a proof. It also requires creativity and critical thinking to come up with a solution, making it a challenging problem for students.

What are some strategies for solving Spivak's Problem?

Some strategies for solving Spivak's Problem include using algebraic manipulation, using the definition of a limit, and using the squeeze theorem. It is also helpful to break the problem down into smaller parts and to think about different approaches to the proof.

How can solving Spivak's Problem help with understanding calculus?

Solving Spivak's Problem can help with understanding calculus because it requires a thorough understanding of the concept of limits, which is a fundamental concept in calculus. It also helps develop critical thinking and problem-solving skills, which are essential in mastering more complex calculus concepts.

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