- #1
bedi
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Prove that the limit of (3n+5)/2(n+1)^2 is 0 when n goes to infinity.
Attempt: I need to find an N such that for any €>0, (3n+5)/2(n+1)^2<€ holds for every n with n>N.
Then I made some manipulation;
(3n+5)/2(n+1)^2 < (3n+5)/(2n^2 +4n) < (3n+6)/(2n^2 +4n) = (n+3)/(n^2 +2n) < (n+3)/n^2
Then what? Please help.
Attempt: I need to find an N such that for any €>0, (3n+5)/2(n+1)^2<€ holds for every n with n>N.
Then I made some manipulation;
(3n+5)/2(n+1)^2 < (3n+5)/(2n^2 +4n) < (3n+6)/(2n^2 +4n) = (n+3)/(n^2 +2n) < (n+3)/n^2
Then what? Please help.