# Proof of False & True Statement: Limit Homework

• MelanieSwan
In summary, when considering the limits of two functions f(x) and g(x) as x approaches infinity, if f(x)< g(x) and both limits exist, then the limit of f(x) is less than the limit of g(x). Additionally, if f(x)<=g(x) for all x in the domain, then the limit of f(x) is less than or equal to the limit of g(x).
MelanieSwan

## Homework Statement

true or false. prove or give counter example:
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

## The Attempt at a Solution

1. I know that this 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.

2. 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.

The inequality 2e + f(x)-g(x) <= 0 is not always true. In particular if f(x)=g(x) it is false.

A better inequality would be
L-M < 2e

And since this holds for any epsilon

L-M<=0.

╔(σ_σ)╝ said:
A better inequality would be
L-M < 2e

And since this holds for any epsilon

L-M<=0.
Could you explain more about this? I don't get the part from L-M < 2e >> L-M<=0 .
Thanks very much for replying :)

MelanieSwan said:
Could you explain more about this? I don't get the part from L-M < 2e >> L-M<=0 .
Thanks very much for replying :)

Okay. :)

f(x) - g(x) < =0

L - M < 2e + f(x) - g(x) <= 2e

Thus by transitivity

L - M < 2e (*)

Since (*) inequality holds for every single epsilon greater than 0 then
L - M <= 0
To see why suppose L - M >0 then L - M = x for some positive real number
if we pick e = x/4 then we get a contradiction. That is,

L - M= x < 2(x/4)

which is a contradiction if x is positive. Note: if x was negative the inequality would be true

Thus L - M <= 0.

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"Proof of False & True Statement: Limit Homework" is a mathematical concept that involves using logical reasoning and evidence to demonstrate the correctness or incorrectness of a statement.

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