Question

determine the limit of :
(3n+4)(1-n)/n^2

Answer 1

when n goes to what

Answer 2

n is raised to the 2nd power, n to infinity

Answer 3

expand the top

Answer 4

-3

Answer 5

the only important terms are 3n*(-n) and n^2

Answer 6

-3n^2/n^2 = -3 for all sufficiently large values of 3 :)

Answer 7

\[\lim_{n \rightarrow \infty}\frac{(3n+4)(1-n)}{n^2}\]
\[\lim_{n \rightarrow \infty}\frac{3n(-n)+\cdots blah}{n^2}\]

Answer 8

-3n^2-n+4/n^2
n^2(-3n-1/n+4/n^2)/n^2=-3n-1/n+4/n^2
when n goes to infinity -1/n+4/n^2goes to 0, only thing left is -3n lim as n goes to infinity -3n=-infinity

Answer 9

gracias