A question about analysis...

Artusartos

1. The problem statement, all variables and given/known data

For question 20.16 (a) in this link:

http://people.ischool.berkeley.edu/~...04hw7sum06.pdf

I don't understand the last sentence in the solution. How/why does the limit comparison test for sequences tell us that result?

Thanks in advance

2. Relevant equations



3. The attempt at a solution

Ray Vickson

Can you have ##f_1(x_n) \leq f_2(x_n) \; \forall n## but ##\lim_{n \to \infty} f_1(x_n) > \lim_{n \to \infty} f_2(x_n)?##

Artusartos

No but isn't that what we are trying to prove?

Of course, when I think about it, it makes sense. But I can't see any theorem like that in my textbook...

Ocifer

I've been thinking a bit about this, and I'm also curious why the Limit Comparison Test should be helpful.

Isn't the limit comparison test related not just to sequences, but specifically to infinite series? Since we're already told that both [itex]f_1[/itex] and [itex]f_2[/itex] converge to finite values as x->a+, why is it helpful that the LCT should tell us they both converge together?

In the proof provided in the OP's link, I follow most of the author's reasoning. I just don't see how LCT comes into it at all.

If this is not rigorous enough, someone please critique, but I am tempted to just leave it at the following:

Let [itex] \langle x_n \rangle [/itex] be a sequence of elements in (a,b) converging to a.

We know:
[itex] f_1(x_n) \leq f_2(x_n)[/itex], for all n

It must follow that:
[itex] \lim_{n \to \infty} f_1(x_n) \leq \lim_{n \to \infty} f_2(x_n)[/itex]

which, by hypothesis, implies:
[itex] L_1 \leq L_2 [/itex]

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Is this also a satisfactory proof?

Ray Vickson

It is probably regarded as obvious; in any case its proof is just about as simple as you can get; just assume the result is false and see what happens.