Question Number 105104 by bemath last updated on 26/Jul/20

lim_(x→0) ((1−cos ^5 (x)cos ^3 (2x)cos ^3 (3x))/(22x^2 ))?

Answered by bramlex last updated on 26/Jul/20

lim_(x→0) ((5cos ^4 (x)sin (x)cos ^3 (2x)cos ^3 (3x)+6cos ^2 (2x)cos ^5 (x)cos ^3 (3x)sin (2x)+9cos ^2 (3x)cos ^5 (x)cos ^3 (2x)sin (3x))/(44x))=  ((5+12+27)/(44)) = 1. ▲

Answered by bobhans last updated on 26/Jul/20

cos x = (√(1−sin ^2 x)) ≈ 1−(x^2 /2)  lim_(x→0) ((1−(1−(x^2 /2))^5 (1−2x^2 )^3 (1−((9x^2 )/2))^3 )/(22x^2 )) =  lim_(x→0) ((1−(1−((5x^2 )/2))(1−6x^2 )(1−((27x^2 )/2)))/(22x^2 )) =  lim_(x→0) ((1−(1−((5/2)+6+((27)/2))x^2 +o(x^2 )))/(22x^2 )) =  lim_(x→0) ((((5/2)+6+((27)/2))x^2 +o(x^2 ))/(22x^2 )) = ((44)/(2(22))) = 1  (B⊚B)