Lim 1-(cosx)(cos2x)(cos3x)(cos4x)/1-cos4x
x-0
2006-07-19 01:56:53 · 8 answers · asked by Parsa D 1 in Science & Mathematics ➔ Mathematics
derivative the function like this-(sinx*cos2x*cos3x*cos4x+sin2x*cosx*cos3x*cos4x+sin3x*cos2x*cosx*cos4x+sin4x*cos2x*cos3x*cosx)/sin4x
(x+2x+3x+4x)/4x=10/4
2006-07-19 02:17:43 · answer #1 · answered by Anonymous · 0⤊ 0⤋
x² = x+6 x² - x - 6 = 0 (x - 3)(x + 2) = 0 So x = 3 or - 2 x² + 2x = 0 x(x + 2) = 0 So x = 0 or - 2 60 = x² - 4x x²- 4x - 60 = 0 (x - 10)(x + 6) = 0 So x = 10 or -6 4x = x² - 40 5 x² - 4x - 40 5 = 0 (x - 9)(x + 5) = 0 So x = 9 or -5 -6x = x² + 9 x² + 6x + 9 = 0 (x + 3)(x + 3) = 0 (x + 3)² = 0 So x = -3 -15=x² + 8x x² + 8x +15 = 0 (x + 5)(x + 3) = 0 So x = -5 or -3 -9x = x² + 14 x² + 9x + 14 = 0 (x + 7)(x + 2) = 0 So x = -7 or - 2 desire that helps. =D
2016-11-06 20:01:46 · answer #2 · answered by Anonymous · 0⤊ 0⤋
cos(ax) ~= 1 - (ax)^2/2 + Cx^4/4 ! - ....
(cosx)(cos2x)(cos3x)(cos4x) ~= 1 - Cx^2 + C2x^4 ...
thus 1 -(cosx)(cos2x)(cos3x)(cos4x) ~= Cx^2 + C2x^4 ...
thus lim for x-> 0 (1 -(cosx)(cos2x)(cos3x)(cos4x)) / x = Cx = 0
2006-07-19 02:26:19 · answer #3 · answered by gjmb1960 7 · 0⤊ 0⤋
cos 0 = 1
As x approaches zero, all of the cosines in your expression approach 1 and the product of multiplying them all together approaches 1, as well.
So you have lim 1 - 1
This is 0.
2006-07-19 03:56:50 · answer #4 · answered by jimbob 6 · 0⤊ 0⤋
0 - cos24x/0 = 0.9135
2006-07-19 02:04:30 · answer #5 · answered by lubetj 1 · 0⤊ 0⤋
answer is 0
2006-07-19 02:00:45 · answer #6 · answered by missy 4 · 0⤊ 0⤋
Complete the equation
2006-07-19 02:30:21 · answer #7 · answered by Dr M 5 · 0⤊ 0⤋
when x->0 then cos(nx)->cos(n*0)=cos(0)=1
1-1*1*1*1....=0
2006-07-19 02:16:07 · answer #8 · answered by mashkas 3 · 0⤊ 0⤋