Find Least and Greatest Bounds of A: (4+X)/X for x≥1 | ε>0 - Solutions

[second]

Homework Statement

A= (4+X)/X WHEN x≥1
Find the least upper bound and greatest lower bound of the following sets.for a given ε>0,
Find a number in the set that exceeds l.u.b. A - ε and a number in the set that is smaller than
g.l.b.A + ε

Homework Equations

does this make sense? is it a right way to attack this question?

The Attempt at a Solution

x≥1 → 1 ≥1/x → 0<1/x≤ 1
is we multiply by 4. →0 < (1/x)4 ≤4
then add one → 1 < (4/x)+1 ≤ 5.
thus 1 is G.l.b of A AND 5 l.u.b of A.

Find a number in the set that exceeds l.u.b. A - ε ...

LET 5 is the upper bound of A. (given)
suppose K is also the upper bound of A. → 5 ≤ K
Suppose this is not the case.AND K < 5.
5-K > 0
ε = 5-K > 0
K =5 -ε SO THAT there is a number X(ε) E A thus x(ε) > 5-ε = K

 

[HallsofIvy]

Yes, 5 is the lub (in fact, it is the maximum value) of A and 1 is the glb.

But I'm not sure what you are doing with:

LET 5 is the upper bound of A. (given)

Well, it's not given- you had to find that 5 is the least upper bound.

suppose K is also the upper bound of A. → 5 ≤ K
Suppose this is not the case.AND K < 5.
5-K > 0
ε = 5-K > 0
K =5 -ε SO THAT there is a number X(ε) E A thus x(ε) > 5-ε = K

You were not asked to prove that there was a number such that A(x) is larger than $5-\epsilon$, you were asked to FIND that number.

For $\epsilon> 0$, suppose that
$$A(x)=\frac{4+ x}{x}= 5- \epsilon$$
then
[tex]4+ x= (5- \epsilon)x[tex]
$$4= (5- \epsilon)x- x= (4- \epsilon)x$$
$$x= \frac{4}{4- \epsilon}$$

 

[second]

would you use the same strategy for G.L.B TOO...WOULDN'T THE NEGATIVE CHANGE THE EQUALITY