what is the limit as n approaches infinity of an=n^3+3n+1/n^2+4n?

Answer 1

If the question is as you have written with four separate terms added together then they can be treated separately. 1/n^2 will tend to zero, the others will tend to infinity. So the sum will tend to infinity.

If the question should have been written (n^3+3n+1)/(n^2+4n) then treat the predominant term and consider the others insignificant.
i.e. as n approaches infinity 3n+1 is insignificant compared to n^3 and 4n is insignificant compared to n^2
the formulae becomes approximately n^3/n^2 = n
so, as n tend to infinity, (n^3+3n+1)/(n^2+4n) also tends to infinity

Answer 2

=n^3(1+3/n^2+1/n^3)/n^2(1+4/n)=n*(1+....)/(1-..)
Each parenthesis =>1 so the limit is +infinite(supposing n >0)

Answer 3

There is no limit, the term n^3 will grow arbirarily large and there isn't anything negative in there to cancel it.