## anonymous 5 years ago lim as n-->infiniti, of a(sub n) = n/(1+(n^1/2)) i do not know how to approach it

1. anonymous

divide each term by n^1/2

2. anonymous

$\lim_{n \rightarrow \infty}= n / 1 + \sqrt{n} solve for convergence or divergence, i do \not know where \to start$

3. anonymous

oh ty

4. anonymous

so it diverges

5. anonymous

i got stuck at diviging n(2/2) / n(1/2)

6. anonymous

we get lim n^(.5) /[ 1/ n^.5 + 1 )

7. anonymous

so thats infinity / ( 0 + 1)

8. anonymous

Or you can realize that the order of the top is n, and the order of the bottom is sqrt(n). Since the order on top is greater, it diverges, so the limit tends to infinity.