1. Dec 16, 2012

### Artusartos

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

For question 20.16 (a) in this link:

http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw7sum06.pdf [Broken]

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

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 6, 2017
2. Dec 16, 2012

### Ray Vickson

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

Last edited by a moderator: May 6, 2017
3. Dec 16, 2012

### 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...

4. Dec 17, 2012

### Ocifer

Isn't the limit comparison test related not just to sequences, but specifically to infinite series? Since we're already told that both $f_1$ and $f_2$ 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 $\langle x_n \rangle$ be a sequence of elements in (a,b) converging to a.

We know:
$f_1(x_n) \leq f_2(x_n)$, for all n

$\lim_{n \to \infty} f_1(x_n) \leq \lim_{n \to \infty} f_2(x_n)$

which, by hypothesis, implies:
$L_1 \leq L_2$

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Is this also a satisfactory proof?

5. Dec 17, 2012

### 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.