what is limit as n--> infinity of n[(n+1)^0.5 - (n^0.5)] ^2
2007-11-20 02:54:26 · 4 answers · asked by dposters 1 in Science & Mathematics ➔ Mathematics
For every n,
n[(n+1)^0.5 - (n^0.5)] ^2 = n[(n^(0.5)(1 + 1/n)^0.5 - (n^0.5)] ^2 = n^2 [(1 + 1/n) - 1]^2.
1/n --> 0 as n --> oo. So, as we know, (1+ 1/n)^0.5 ~ 1+ 1/n * 0.5 = 1 + 1/(2n). It follows that
lim n--> infinity of n[(n+1)^0.5 - (n^0.5)] ^2 = lim n --> oo n^2[1 + 1/(2n) -1 ]^2 = lim n --> oo n^2 [1/(2n)]^2 = lim n --> oo n^2 * 1/(4n^2) = 1/4 = 0.25
This does NOT go to oo because (n+1)^0.5 - (n^0.5) --> 0 as n --> oo.
2007-11-20 03:08:17 · answer #1 · answered by Steiner 7 · 1⤊ 0⤋
i'm not sure of the solution, but i'm placing my bets to 0. as the difference of the two square roots would approximate 1/n, and given it is squared and multipled to n, we get 1/n, which would approximate to zero. i'm still trying to prove that (n+1)^0.5-n^0.5 would actually approximate to 1/n.
p.s. you can try this with your scientific calculator. set n as 9999999999999 and apply to the equation. i got an answer of 0.25.
2007-11-20 11:12:25 · answer #2 · answered by hapeehadee 2 · 0⤊ 1⤋
limit n--> infinity of n[(n+1)^0.5 - (n^0.5)] ^2
Since, all the terms of (n) is in the numerator,
there is no chance of them being canceled out.
Thus, the limit of the above is infinity.
2007-11-20 11:07:54 · answer #3 · answered by crashbird 2 · 0⤊ 1⤋
Take the derivative equals zero.
2007-11-20 11:01:41 · answer #4 · answered by Shary 6 · 0⤊ 1⤋