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Well there is this problem that i am struggling to proof, i think i am close but nope nothing yet. Well, the problem goes like this:

We have two sequences [tex] \ {(a_n)} [/tex] and [tex] ({b_n})[/tex], whith the feature that [tex]{a_n}<{b_n}[/tex] for every n. Also, [tex]{a_2}={b_1}, {a_3}={b_2}... [/tex] and so forth. What i need to show is that they have the same limit, that is:

[tex]\lim_{n\rightarrow infinit} {a_n}=m=\lim_{n\rightarrow infinit} {b_n}[/tex]

I started by using the definition of the limit like this,

First i assumed that the limits are not the same so i put:

[tex]\lim_{n\rightarrow infinit} {a_n}=a[/tex] and [tex]\lim_{n\rightarrow infinit} {b_n}=b [/tex], where a is not equal to b, so it must be either smaller or bigger than it.

For every epsilon, there exists some number N_1, that for every n>N it is valid that:

I a_n-a I< epsylon, and also for every epsylon there must exist some number call it N_2, that for every n>N_2 we have:

I b_n- b I< epsylon,

then i took N=max{N_1,N_2} so for n>N, both

I b_n- b I< epsylon and I a_n-a I< epsylon are valid

then after some transformations i got:

a- epsylon<a_n< a+ epsylon and

b-epsylon< b_n < b+ epsylon

let us now suppose that b<a, than this is not possible since from a theorem i could find a number call it n_o that for every n>n_o also b_n<b<a_n , which actually contradicts the very first supposition that b_n>a_n.

now let us take b>a, and also from the supposition that b_n>a_n whe have

a-epsylon<a_n<b_n<b+epsylon

and this is where i get stuck.

Any suggestions would be really appreciated.

, so from here we have that the limit of a_n would be both b and

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# Sequences, need help

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