Convergence and limits analysis problem

[dancergirlie]

Homework Statement

Assume that (xn) and (yn) are sequences, both of which
converge to L. Assume further that (zn) is a sequence satisfying
xn <or= zn <or= yn

for all n in N. Prove that (zn) also converges to L

Homework Equations

The Attempt at a Solution

Let xn and yn be sequences that converge to L and let zn converge to L'
The fact that xn is <or= to zn for all n in N implies that L<or=L'
The fact that zn is <or= to yn for all n in N implies that L'<or=L'
Meaning L<or=L'<or=L
Therefore, L=L'

Would this be correct? Do I need to show that Zn converges before I assume it has a limit?

 

[slider142]

Let xn and yn be sequences that converge to L and let zn converge to L'
The fact that xn is <or= to zn for all n in N implies that L<or=L'
The fact that zn is <or= to yn for all n in N implies that L'<or=L'
Meaning L<or=L'<or=L
Therefore, L=L'

Would this be correct? Do I need to show that Zn converges before I assume it has a limit?

Yes. You have not shown that zn converges at all; we can't start by assuming it converges, unless we later plan to append a proof for the case where it is not assumed to converge, in which case the latter proof is stronger and replaces the former!
Also, I'm not sure what theorems you have already proven, but you may need to show the proof for the second line's inequality limit implication.

 

[dancergirlie]

we've already proven the theorem that if
lima<or= limb for all n in N then a<b so I assumed I could just use that theorem.

How would you recommend I go about proving that zn converges? Should I do an induction proof? Because the sequence is bounded, so all I'd have to prove would be that it is either increasing or decreasing, right?

 

[dancergirlie]

i'm sorry i mean we have already proven that
if lima<or=limb for all n in N then a<or=b

 

[slider142]

i'm sorry i mean we have already proven that
if lima<or=limb for all n in N then a<or=b

Unfortunately, the existence of lim b is part of the assumption for that theorem, so it cannot be used directly until we show zn does indeed converge. This is actually easier than it looks and need only uses the definition of the limit of a sequence.
Consider the two inequalities:
xn - L <= zn - L <= yn - L
L - xn >= L - zn >= L - yn
which are perfectly valid pointwise. Then add in the definition of the limit of the sequences x and y tending to L, and show that this gives the definition of z converging to L as well.

 

[dancergirlie]

Alright how does this look?

assume xn<=zn<=yn
Meaning, xn-L<=zn-L<=yn-L

Since L is the limit of xn and yn, that means:
for e>0 there exists N1,N2 in N so that:

|xn-L|<e and |yn-L|<e
which is equivalent to:
-e<xn-L<e and -e<yn-L<e
since -e<xn-L and yn-L<e
that means:
-e<xn-L<=zn-L<=yn-L<e
and -e<zn-L<e
meaning for e>0 there exists an N in N so that
|zn-L|<e
and therefore, L is the limit of zn

 

[slider142]

Excellent work. Very straightforward.

 

[dancergirlie]

thanks so much for the help :D