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#
