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

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Answer 1 · 2017-06-05T09:15:50

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:

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

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