Answer:
a) [tex]\lim_{n\rightarrow \infty}a_n=0[/tex]
b) The limit is [[4/3]]=1, that is, [tex]\lim_{n\rightarrow \infty}a_n=1[/tex]
Step-by-step explanation:
The definition of the limit of a sequence (an) is as follows:
[tex]\lim_{n\rightarrow \infty}a_n=L[/tex] (the sequence (an) converges to L) means that given [tex]\epsilon>0[/tex], there exists some natural number N such that if n≥N, then [tex]|a_n-L|<\epsilon[/tex]
We will use that [[x]] is a increasing function. That is, if x≤y then [[x]]≤[[y]]
a) For all [tex]\epsilon>0[/tex], take N≥6. Then 1/N ≤1/6 and 5/N ≤ 5/6 < 1. We have that 0<5/N<1, therefore [[5/N]]=0 (0 is the biggest integer less than or equal than x, for any x on the interval 0<x<1)
For n≥N, we have that 1/n ≤ 1/N, hence 0< 5/n ≤ 5/N < 1 . As we discussed above, [[5/n]]=0. Then |a_n|=0, therefore [tex]|a_n-0|=|a_n|=0<\epsilon[/tex]
b) For all [tex]\epsilon>0[/tex], take N≥3/4 (any N will do). Then 1/N ≤ 4/3 and 1/4N ≤ 1/3. Note that a_n=[[(12 + 4n)/3n]]=[[1/4n + 4/3]]
For n≥N, we have that 1/4n ≤ 1/4N, hence 1< 4/3< 1/4n +4/3 ≤ 1/4N+4/3 ≤ 5/3 <2.If 1<x<2 then [[x]]=1 , hence a_n=[[1/4n +4/3]]=1
Then |a_n-1|=|1-1|=0, therefore [tex]|a_n-1|=0<\epsilon[/tex]
