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#