Question

How do you prove this ? if `lim_(n->oo)x_(2n)=L=lim_(n->oo)x_(2n+1)` then `lim_(n->oo)x_(n)=L`

Answer 1

Start from:

`lim_(n->oo) x_(2n) = L`

This means that for any number `epsilon > 0` we can find an `N_e in NN` such that, for `2n > N_e`

`(1)abs(L-x_(2n)) < epsilon` for `2n > N_e`

Similarly:

`lim_(n->oo) x_(2n+1) = L`

implies that or any number `epsilon > 0` we can find an `N_o in NN` such that, for `2n+1 > N_o`

`(2) abs(L-x_(2n+1)) < epsilon` for `2n+1 > N_o`

Then for any number `epsilon > 0`, if we choose `N in NN` such that `N >= max(N_e,N_o)`, then we have that for every `n > N`

`(3) abs(L-x_n) < epsilon` for `n > N`

based on (1) if `n` is even and on (2) if `n` is odd, which proves that:

`lim_(n->oo) x_n = L`