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 .
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.
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?
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
I don't think that's what a formal argument means.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
This is not true, you are making the same mistake.>> -2e < f(x) - g(x) + L - M < 2e since 2e > 0
>> -2e < f(x) - g(x) + L - M < 0
What I am trying to suggest is you start at the place you had , assume that and show that this leads to a contradiction.