# More limits

1. Jul 16, 2007

### daniel_i_l

1. The problem statement, all variables and given/known data
Prove or disprove:
You have the functions g:R->R and f:R->R:
1) If the limit of g at x0 is infinity and the limit of f*g (multiplication) at x0 is also infinity then there's some neibourhood of x0 where f(x)>0 for every x in the neibourhood.

2)f and g are defined only in [0,infinity) and and L is in R:
If the limit of f*g at infinity is L and the limit of f at infinity is infinity then the limit of g at infinity exists.

2. Relevant equations

3. The attempt at a solution
1) False: g(x) = 1/(x^2) and f(x) = 1 for all x=/=0 and f(x) = -1 for x=0.
Is that right? It seemed to trivial.
2) True: For every E (epsilon) >0 we can find M>0 so that ME>|L| =>
ME-|L| > 0.
Also, we can find N_1>0 so that for all x>N_1 f(x)>M => |f(x)| > M.
Also, we can find N_2>0 so that for all x>N_2 |f(x)g(x)-L|< ME-|L| .
So if N>max(N_1,N_2) then:
ME-|L| < |f(x)g(x)-L|<= |f(x)g(x)|-|L| < M|g(x)|-|L| =>
|g(x)| < E
and so we found that the limit of g at infinity is 0. Is that right? It seems weird that in the question they only proved that the limit exists and I found that it's always 0.

Thanks.

2. Jul 17, 2007

### daniel_i_l

Anyone? An in the line:
ME-|L| < |f(x)g(x)-L|<= |f(x)g(x)|-|L| < M|g(x)|-|L|
I meant:
ME-|L| > |f(x)g(x)-L| => |f(x)g(x)|-|L| > M|g(x)|-|L|

3. Jul 17, 2007

### StatusX

I think you're example for the first one might be too trivial. Since the value of a limit of a function never depends on the value at the limit point, they probably don't mean to include the point in the neighborhood. If that's exactly what they wrote, then you're right, but it's a little cheap.

And I think your result for the second question is correct, although I'm having trouble following your proof. For example, ME>|L| => ME-|L| > 0 is always trivially true, and there are a few other examples like that where I don't know what you mean.

4. Jul 17, 2007

### daniel_i_l

Where else was I unclear?
Thanks!

5. Jul 17, 2007

### StatusX

As I said, this is trivially true, so what exactly is the condition you're imposing on M?

This is a little confusing, although I'm guessing you mean x>N_1 => f(x)>M => |f(x)|>m.

Without knowing M I don't know what this means.

|A-B|<= |A|-|B| is not true unless you establish |A|>=|B|, in which case it's an equality.

6. Jul 17, 2007

### daniel_i_l

I'm imposing the condition that ME > |L|. Since E can be every number over 0 this isn't true for all M.

yes.

Since the limit of f*g at infinity is L, then for every a>0 we can find a number N>0 so that |f(x)g(x) - L|<a and since we chose M so that ME-|L|>0 we can find a number N>0 so that |f(x)g(x) - L|<ME-|L|

Look at my second post, in the first one I mixed up all the signs in that row.

Is it clear now? Is it a correct/complete proof?
Thanks

7. Jul 17, 2007

### StatusX

Ok, now I'm with you up to the last line:

ok

does this mean "greater than or equal to" or "implies"? If it's the first, the comment I made above still applies, and if it's the second, I don't know what you mean.

In any case, it's still confusing, at least to me. Just for the record, here's how I would do it:

There is some xo such that for x>xo, |f(x)|>M and |f(x)g(x)-L| is less than, say, 1. Then for x>xo we can write:

$$|g(x)| = \left| \frac{ (f(x)g(x) - L)+L}{f(x)} \right|$$

$$\leq \frac{1}{|f(x)|} \left( |f(x)g(x)-L| + |L| \right)$$

$$< \frac{1+|L|}{M}$$

Last edited: Jul 17, 2007
8. Jul 17, 2007

### daniel_i_l

Your way does seem much simpler. Do you have any advice for how to see those kind of "tricks"? (the first equation you latexed)
"does this mean "greater than or equal to" or "implies"? If it's the first, the comment I made above still applies, and if it's the second, I don't know what you mean."
I meant greater than or equal to. And isn't it true that for all a and b
|a-b|>=|a|-|b| ? proof:
|a| = |a-b+b| =< |a-b|+|b|
and we get:
|a-b| >= |a|-|b|

Is that the problem? If not than what's the problem?
Thanks.

9. Jul 17, 2007

### StatusX

Sorry, you're right. Your way works fine. The only real difference between my way and yours is that you used the fact that f(x)g(x) approaches L. But this is actually unneceassary, and it simplifies the proof somewhat if you note that f(x)g(x) just needs to be bounded for the result to hold.

10. Jul 17, 2007

### daniel_i_l

Thanks a lot for your help!