Question

Sa se calculeze limita:
a)(2n^3-n^2+1)
b) (3^n+1)/ (2^n+1)

Answer 1

a) 2n^3-n^2+1=n^2(2n-1)+1 ->(n->inf) inf*inf+1=inf.
b) (3^n+1)/(2^n+1)=((3/2)^n+1/2^n)/(1+1/2^n) ->(n->inf) (inf+0)/(1+0)=inf.