Question

Find if divergent/convergent:
a(sub n)=(n^2/sqr(n^3+4n)), if convergent, find the limit.

Answer 1

where are you stuck??

Answer 2

how to start it <<<

Answer 3

what tests do you know?

Answer 4

squeeze theorm, using f(x) and take the limit

Answer 5

this is what i tired to do.... f(x)=(x^2(/sqr(x^3+4x)), and take the limit, but dont know how

Answer 6

\[\sqrt{n^3+4n}=\sqrt{n^3\left(1+\frac{4}{n^2}\right)}=n^{3/2}\sqrt{1+\frac{4}{n^2}}\]

Answer 7

yes, i divided top/bottom by n^3, and got (1/n)/(sqr(1+(4/(n^2))))

Answer 8

so it divergent because top is zero when take the limit?

Answer 9

try that again...you didn't do it correctly...use what I wrote

Answer 10

sqr(n)/((sqr(1+4/(n^2))) ??

Answer 11

yes

Answer 12

so limit as x>inifnity, is infinty/1 ?,, that doesnt seem right

Answer 13

so the top is going to infinity...the bottom to 1...that means that the limit is infinity (or does not exist..depending on what the teacher wants)

Answer 14

i c... thanks, the book says Divergent, so it is infinity/not exist.

Answer 15

same thing

Answer 16

kk, ty for the help