how do you find the limits of his?
lim x^9 cos (3/x)
x => 0
2007-10-20 06:25:26 · 3 answers · asked by remote control 1 in Science & Mathematics ➔ Mathematics
Note that since |cos (3/x)| ≤ 1, regardless of the value of x. Therefore, we have the inequality - |x^9| ≤ - |x^9| |cos (3/x)| ≤ x^9 cos (3/x) ≤ |x^9| |cos (3/x)| ≤ |x^9| everywhere except 0, where cos (3/x) is undefined. Since [x→0]lim -|x^9| = [x→0]lim |x^9| = 0, it follows from the squeeze theorem that [x→0]lim x^9 cos (3/x) = 0. Q.E.D.
2007-10-20 06:40:45 · answer #1 · answered by Pascal 7 · 0⤊ 0⤋
as -1 <=cos(3/x)<=1 and x^9==>0 the limit is 0
2007-10-20 13:37:31 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
For limits you plug in what x is approaching
x =>0
so (0)^2 * cos(3/0) ===== 0
Plain and simple
2007-10-20 15:27:34 · answer #3 · answered by Anonymous · 0⤊ 0⤋