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