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

• Bleys
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.
Bleys
I came across a problem in Calculus by Spivak, and I'm having trouble formalizing the proof.
Let $$A_{n}$$ bet a set of finite numbers in [0,1], and if $$m \neq n$$ then $$A_{n}$$ and $$A_{m}$$ are disjoint. Let f(x) be defined as
f(x)=1/n if x is in $$A_{n}$$ and f(x)=0 if x is not in any $$A_{n}$$. 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 $$A_{n}$$ in a neighborhood of a. Choose the smallest such n, say $$n_{0}$$. Then $$f(x)\leq1/n_{0}$$. Restrict the neighborhood further so that none of the elements of $$A_{n_{0}}$$ are in the interval. Then choose the next n such that it's minimal. Obviously $$f(x)\leq1/n\leq1/n_{0}$$. 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 $$\delta$$ to choose?

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 $$\varepsilon>0$$, then there exists N > 0 with $$1/N < \varepsilon$$. Let $$B_N = \cup_{ n = 1 }^N A_n$$, so B_N is a finite set, and $$a \not \in B_N$$ (why?). Now, let $$\delta > 0$$ be such that $$( a - \delta, a + \delta ) \cap B_N = \emptyset$$ (why does such a $$\delta$$ exist?). Then if $$| a - x | < \delta$$, either $$x \in A_m$$ for m > N, or f(x) = 0. In either case, $$|f(a) - f(x)|<1/N<\varepsilon$$, as desired.

Bleys said:
I came across a problem in Calculus by Spivak, and I'm having trouble formalizing the proof.
Let $$A_{n}$$ bet a set of finite numbers in [0,1], and if $$m \neq n$$ then $$A_{n}$$ and $$A_{m}$$ are disjoint. Let f(x) be defined as
f(x)=1/n if x is in $$A_{n}$$ and f(x)=0 if x is not in any $$A_{n}$$. 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 $$A_{n}$$ in a neighborhood of a. Choose the smallest such n, say $$n_{0}$$. Then $$f(x)\leq1/n_{0}$$. Restrict the neighborhood further so that none of the elements of $$A_{n_{0}}$$ are in the interval. Then choose the next n such that it's minimal. Obviously $$f(x)\leq1/n\leq1/n_{0}$$. 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 $$\delta$$ 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.

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 $$A_{n}$$ 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 $$\delta$$ exist?)
Because B_N is finite.

Did you assume f(a)=0 because you had $$( a - \delta, a + \delta )$$? Would $$0 < | a - x | < \delta$$ solve that? After all, since we are looking at the limit 0 then the expression $$|f(a) - f(x)|<\varepsilon$$ becomes $$|f(x) - 0|=|f(x)|<\varepsilon$$, 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 $$1/n<\varepsilon$$) 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 $$0 < | a - x | < \delta$$ would solve that.
Ok I think I get it. I was having trouble formulating all this with the $$\delta - \varepsilon$$ definition. Thanks for all your help!

Bleys said:
Sorry, what I meant to say was the smallest n such that the set $$A_{n}$$ 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 $$( a - \delta, a + \delta )$$? Would $$0 < | a - x | < \delta$$ solve that? After all, since we are looking at the limit 0 then the expression $$|f(a) - f(x)|<\varepsilon$$ becomes $$|f(x) - 0|=|f(x)|<\varepsilon$$, since the value of f at a doesn't matter.So since S will be finite (since there will be an n such that $$1/n<\varepsilon$$) 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 $$0 < | a - x | < \delta$$ would solve that.
Ok I think I get it. I was having trouble formulating all this with the $$\delta - \varepsilon$$ 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 $$\in$$ $$\cup$$ 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

## 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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