Spivak Calculus Problem 5-23

[lovebear123]

Spivak "Calculus" Problem 5-23

Spivak "Calculus" Problem 5-23: For your convenience, here is the problem:

Prove that if neither of the following two holds :
1. lim x ->0 f(x) exists and is not 0
2. lim x ->0 |f(x)| = infinity

then there is a function g such that lim x ->0 g(x) does not exist, but lim x ->0 f(x)g(x) does exist. Hint: Consider separately the following two cases : (1) For some e>0, we have |f(x)| > e for all sufficiently small x. (2) For every e>0, there are arbitrarily small x with |f(x)| < e. In the second case, begin by choosing points Xn with |Xn| < 1/n and |f(Xn)| < 1/n.

Note: e means epsilon
My attempt:

Case 1: for some e > 0, as long as x is small enough, we have |f(x)| > e. Let g = 1/f, obviously this does not hold for all e>0 (otherwise
lim x ->0 |f(x)| = infinity). so there is some e1 such that the case title does not hold, there are arbitrarily small x with |f(x)| < e1, for
e' = 1/e1>0, there are arbitrarily small x with |g(x)| = |1/f(x)| > e' = 1/e1. so lim x ->0 g(x) does not exist, and lim x ->0 f(x)g(x) =1

Case 2:for every e>0, no matter how small x is restricted to, there are some x such that |f(x)| < e. Let g = 1/f, for each N>0, according to the assumption, we have for e = 1/N, there are arbitrarily small x with |x| < 1/N and |f(x)| < e = 1/N, so we have arbitrarily small x with |g(x)| > N, so lim x->0 g(x) does not exist and lim x ->0 f(x)g(x) =1

My Questions are:
1. I am totally not sure about my attempt
2. I am confused by the answer given in the answer book (by Spivak), especially concerning Case 2, Spivak writes:
In case 2, choose Xn as in the hint. Define g(x) = 0 for x is not Xn, and g(x) =1 for x= Xn. Then lim x -> 0 g(x) does not exist, but
lim x ->0 f(x)g(x) = 0.

I totally don't understand this, for example, f(x) = sin(1/x), when n =1, X1<1, f(X1)<1, obviously all x satisfying |x| <1 would be suitable Xn (when n = 1), but obviously g(x) =1 for all such x won't make
lim x ->0 f(x)g(x) = 0.

Actually it's a total mess, I don't understand the whole problem at all, I don't know the intention for such two cases...Need help~

I really hope that someone could help me with this, and this is my first experience in this physicsforums, so please... Thank you very much!

 

[mighty2000]

Thank you for sharing your attempt and questions regarding Spivak's "Calculus" Problem 5-23. Here is my explanation and clarification on the problem and the solution provided in the answer book.

First, let's understand what the problem is asking us to prove. We are given two conditions, neither of which holds:

1. lim x ->0 f(x) exists and is not 0
2. lim x ->0 |f(x)| = infinity

From these conditions, we need to prove that there exists a function g such that lim x ->0 g(x) does not exist, but lim x ->0 f(x)g(x) does exist. This means that we need to find a function g that does not have a limit at x = 0, but when multiplied with f(x), the resulting function has a limit at x = 0.

Now, let's look at the two cases provided in the problem:

Case 1: For some e > 0, as long as x is small enough, we have |f(x)| > e.

This means that for any positive value of e, there exists a small enough value of x for which |f(x)| is greater than e. In this case, we can choose g = 1/f. This function does not have a limit at x = 0 because f(x) does not have a limit at x = 0. However, when we multiply f(x) with g(x), we get a constant function with a limit of 1 at x = 0.

Case 2: For every e > 0, there are arbitrarily small x with |f(x)| < e.

This means that for any positive value of e, we can find a small enough value of x for which |f(x)| is less than e. In this case, we need to choose Xn as given in the hint, i.e. Xn = 1/n for all positive integers n. Now, we define g(x) as follows:

g(x) = 0 for x ≠ Xn
g(Xn) = 1 for x = Xn

This function g does not have a limit at x = 0 because for any small enough value of x, g(x) will either be 0 or 1. However, when we multiply f(x) with g(x), we get a function that is 0 for all x ≠