# Thread: function with greater value has greater limit?

1. ## function with greater value has greater limit?

I'm a little bit confused about these similar problems in my analysis text book, and I hope to get some little help. It says, true or false:

1. with x approaches to infinity suppose that lim f(x) = L and lim g(x) = M and f(x)<= g(x), then L<=M
2. with x approaches to infinity suppose that lim f(x) = L and lim g(x) = M and f(x)< g(x), then L< M.
for all x in D
I know that the second one is wrong and one counter example I can think of is: f(x) = 2/|x| and g(x) = 3/|x| so for all non zero x, f(x) < g(x) but their limits are both zero. Is it just simple as that or I miss something?

Also for the first one, I think it might be wrong but I cannot think of a counter example. I'm much appreciated if anyone could give me some hints. Thanks in advance.

2. I think you are right about the second proposition.

The first one is true. To see that, you only need to use the definition of limit and the condition that $f(x)\leq g(x)$.

3. I came up with this. I wonder if it's correct?
for every e > 0 and e is very very small then there exists m1 in N such that for every x>= m1 we have: |f(x) - L| <e rewrite as e> f(x) - L> -e (1)
similarly there exists m2 in N such that for every x>=m2 we have |g(x) - M| <e rewrite as -e< g(x)-M <e (2)
subtract (1) from (2) we have:
-2e < g(x) - f(x) + L- M) < 2e.
we only care about the right part of the inequation above so:
L-M < 2e +f(x) -g(x)
since f(x) - g(x) <= 0, there always exists e positive but very very small so that 2e + f(x) - g(x) <= 0
so L-M <= 0
so L <= M as desired.

Do you think i got it wrong somewhere?

4. I think you are close to the solution, but there are a few problems with your argument. You say that $f(x)-g(x)\leq0\Rightarrow 2\varepsilon+f(x)-g(x)\leq 0$ which is not true ( $2\varepsilon>0$!) and even if it was
$L-M<2\varepsilon+f(x)-g(x)\leq 0\Rightarrow L-M\leq0$ is not.
This requires more careful treatment. An easy way I see involves using the method of contradiction.

5. thanks very much for your response. contradiction? I wonder if this works?

from L-M < 2e +f(x) -g(x) and f(x) - g(x) <= 0
>> L-M < 2e + f(x) -g(x) <= 2e
>> L-M < 2e

but it must be for all e > 0 so
L-M <= 0
>> L <= M

6. Originally Posted by EmmWalfer
L-M < 2e

but it must be for all e > 0 so
L-M <= 0
Can you prove formally your statement?
By using contradiction I meant this: let's assume that $L>M$. What then, can you finish?

7. L-M < 2e and this must be for ALL e>0 including some e very very small and very very close to 0. In order for this to be always true in case e gets really really small, and really really close to 0, L-M must be be less than or equal to 0

so L <= M

I don't know how to work it out with the contradiction as you said:

if L> M but f(x) <= g(x) so f(x) - L < g(x) - M >> f(x) - g(x) - L + M < 0
but by definition: | f(x) - L| < e
|g(x) - M| < e
>> |f(x) - L - M + g(x)| < 2e

>> -2e < f(x) - g(x) + L - M < 2e since 2e > 0
>> -2e < f(x) - g(x) + L - M < 0
>> 0 < f(x) - g(x) + L - M + 2e < 2e
when f(x) = g(x) >> L-M + 2e < 2e
but L-M > 0 since L >M >> contradiction

so L<= M.

That's all i could think of using contradiction

8. L-M < 2e and this must be for ALL e>0 including some e very very small and very very close to 0. In order for this to be always true in case e gets really really small, and really really close to 0, L-M must be be less than or equal to 0

so L <= M
I don't think that's what a formal argument means.

>> -2e < f(x) - g(x) + L - M < 2e since 2e > 0
>> -2e < f(x) - g(x) + L - M < 0
This is not true, you are making the same mistake.

What I am trying to suggest is you start at the place you had $L-M<2\varepsilon$, assume that $L>M$ and show that this leads to a contradiction.

9. ok I tried again. Thanks for being patient with me:

L>M >> L-M> 0
>> -2e < 0 < L-M < 2e >> |L-M| < 2e but if so then L = M so contradiction!

???

10. No problem. I think it's ok now. For the sake of completeness you should verify why your last statement is true though.

My idea was to do it like this:
$L>M\Rightarrow\exists \delta>0:\ L=M+\delta\Rightarrow \frac{1}{2}(L-M)=\frac{1}{2}\delta<\varepsilon, \forall \varepsilon>0$. This is a contradiction since you can choose $\varepsilon=\frac{1}{3}\delta<\frac{1}{2}\delta$.