Question

lim_(n->infinity) (n+1)^(n+1)/((n+1)^(n+1)-n^n) = 1 ... this is what wolfram says.. could anyone explain me the steps? .. http://www.wolframalpha.com/input/?i=%28n%2B1%29^%28n%2B1%29+%2F+%28%28n%2B1%29^%28n%2B1%29+-+n^n%29

Answer 1

Divide numerator and denominator by \[ (n+1)^{n+1} \] and then you should be able to show both go to 1 as n goes to infinity.

Answer 2

sorry, i worked like 50 limits and started to get tired.. could not see that distribution.. thanks .

Answer 3

\[\lim_{n \rightarrow \infty}\]
(n+1)^n+1
÷
(n+1)^n+1-n^n=
\[\lim_{n \rightarrow \infty}\]
(n+1)^n+1
÷
(n+1)-n *
\[\lim_{n \rightarrow \infty}\]
(n+1)-n
÷
(n+1)^n+1-n^n