Limit proof


by Bleys
Tags: limit, proof
Bleys
Bleys is offline
#1
Nov27-09, 09:22 AM
P: 74
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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rochfor1
rochfor1 is offline
#2
Nov27-09, 10:02 AM
P: 257
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.
wisvuze
wisvuze is offline
#3
Nov27-09, 10:31 AM
P: 367
Quote Quote by Bleys View Post
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.

Bleys
Bleys is offline
#4
Nov27-09, 12:40 PM
P: 74

Limit proof


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!
wisvuze
wisvuze is offline
#5
Nov27-09, 01:05 PM
P: 367
Quote Quote by Bleys View Post
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


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