anonymous
  • anonymous
lim as n-->infiniti, of a(sub n) = n/(1+(n^1/2)) i do not know how to approach it
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
divide each term by n^1/2
anonymous
  • anonymous
\[\lim_{n \rightarrow \infty}= n / 1 + \sqrt{n} solve for convergence or divergence, i do \not know where \to start\]
anonymous
  • anonymous
oh ty

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so it diverges
anonymous
  • anonymous
i got stuck at diviging n(2/2) / n(1/2)
anonymous
  • anonymous
we get lim n^(.5) /[ 1/ n^.5 + 1 )
anonymous
  • anonymous
so thats infinity / ( 0 + 1)
anonymous
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.