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

1 Answer
Jun 5, 2017

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