where n~2=n*n;
2007-11-23 22:45:31 · 4 answers · asked by anacond14 2 in Science & Mathematics ➔ Mathematics
limit of sin(PI*sqrt(n~2+7*n+1))?? when n->infinite
n~2+7*n+1 -> infinity when n-> infinity
but
the sin function oscillates between +1 and -1 indefinitely and has NO limit at infinity
2007-11-23 23:23:29 · answer #1 · answered by vlee1225 6 · 1⤊ 0⤋
sin(sqrt(n^2+7n+1)-n)pi =
(-1)^n sin (7n+1)/[sqrt(n^2+7n+1)+n]*pi
sin (7n+1)/[sqrt(n^2+7n+1)+n]===> sin7/2pi = -1
so the expression oscillates between 1 an -1
Explanation
If you subtract n*pi from an angle sin does not change if n is even and changes it sign if n is odd
2007-11-23 23:54:11 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
sin(PI*sqrt(n^2+7*n+1))->sin(PI*n) =0 when n->infinite
2007-11-23 23:17:21 · answer #3 · answered by Vlad N 2 · 0⤊ 1⤋
the answer is indeterminate
2007-11-23 23:52:26 · answer #4 · answered by mikeasieni 1 · 0⤊ 0⤋