How do you prove if $lim_(n->infty) x_(2n) = L = lim_(n->infty) x_(2n+1)$ , then $lim_(n->infty) x_n = L$ ?

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Answer 1 · 2017-07-27T21:22:26

If $lim_(n->oo) x_(2n) = L$ , this means that for any number $epsilon_e > 0$ we can find $N_e$ such that:

$n > N_e => abs(x_(2n)-L) < epsilon_e$

On the other hand, $lim_(n->oo) x_(2n+1) = L$ means that for any number $epsilon_o > 0$ we can find $N_o$ such that:

$n > N_o => abs(x_(2n+1)-L) < epsilon_o$

Choose now: $epsilon = min(epsilon_e, epsilon_o)$ and $N = max(N_e,N_o)$

Clearly for any $n > N$ we have that if $n$ is even, as $n > N >= N_e$ :

$abs(x_(n)-L) < epsilon_e <= epsilon$

and if $n$ is odd, as $n > N >= N_o$ :

$abs(x_(n)-L) < epsilon_o <= epsilon$

Either way:

$n > N => abs(x_(n)-L) < epsilon$

which proves that:

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

How do you prove if lim_(n->infty) x_(2n) = L = lim_(n->infty) x_(2n+1)lim_(n->infty) x_(2n) = L = lim_(n->infty) x_(2n+1), then lim_(n->infty) x_n = Llim_(n->infty) x_n = L?