Rei Andros wrote:NZG wrote:
Find the limit of the sequence as n goes to infinity
n=(ln(n+1)^2)/(n^2)
There are no typos and best thing is.
No answer for it..
lim n->inf (ln(n+1)^2)/(n^2)
It's been a long time but I think you should apply L'Hospital's Rule. If you plug in the limit you get what's called infinity/infinity form. L'Hospital's Rule allows you to take the derivative of the function in this case.
By taking the derivative you get:
lim n->inf (ln(n+1)^2)/(n^2) = [2(n+1)]/(n+1)^2/2n
Simplify
lim n->inf (ln(n+1)^2)/(n^2) = 1/(n^2+1)
Divide the top and bottom by n^2
lim n->inf (ln(n+1)^2)/(n^2) = 1/n^2/(1+1/n^2)
Now plug in the limit again. Both 1/n^2 terms go to zero so the answer should be zero.
Groce
Whoever canceled this how has the official title of biggest douchebag on Earth.
But it's all calculus stuff, no? What are you doing now?
sooooooooooo
