prove that n/(n^2 + 3) --> 0 as n --> infinity?

Answer 1

Be ε a positive number, let's find a positive number X such that
n>X -> n/(n^2 + 3) < ε

n/(n^2 + 3) < n/n^2 = 1/n

Take X = 1/ε

Answer 2

Factor out an n from the bottom:
n / n(1n+3/n)
= 1/(1n + 3/n) and 1n goes to infinity and 3/n goes to zero as n goes to infinity. So the denominator goes to infinity + 1. So the entire quantity goes to 0 as n goes to infinity.