use the definition of limit of a sequence toshow that limit (3n+1)/(2n^2+3)=0 as n tends to infinity?

Answer 1

abs((3n+1)/(2n^2+3)) = abs((3/n+1/n^2)/(2+3/n^2)) < abs(3/n+1/n^2) < abs(3/n) = 3/n < epsilon if n>3/epsilon,

so, for every epsilon>0, there exists an N, possibly depending on epsilon (in this case N=3/epsilon), such that
abs(an-0)N.

Answer 2

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

= (3+1/n)/(2n+3/n)
= (3 + 0) / (2n + 0) [1/n → 0]
= 3 / 2n
= 3/2 * 1/n
= 3/2 * 0
=0

Answer 3

the priority is to locate the decrease of (2^x)/ (x^2) as x has a tendency to ? (I have replaced your variable 'n' to 'x' for convenience. the following as both the numerator and denominator have a tendency to ? we word L'clinical institution's rule {Lt (x -> ?) f(x)/g(x) = Lt (x -> ?) f '(x)/g'(x)} to substantiate the certainly decrease and we can see that we ought to computer screen it two times. The decrease after 2 software of L'clinical institution's rule reduces to Lt (x -> ?) (2^x)/ (x^2) = Lt (x -> ?) {(ln2*ln2/2)*2^x} = ? subsequently the function has a tendency to Infinity.