Question Number 104899 by bramlex last updated on 24/Jul/20

lim_(x→0) ((cos (sin x)−cos (x))/x^4 ) ?

Answered by john santu last updated on 24/Jul/20

lim_(x→0)  ((−2sin (((x+sin x)/2))sin (((sin x−x)/2)))/x^4 )  lim_(x→0) ((−2(((x+sin x)/2))(((sin x−x)/2)))/x^4 )  lim_(x→0) −(((x+x+(x^3 /(3!)))(x+(x^3 /(3!))−x))/(2x^4 ))  lim_(x→0)  −(((x^3 /6)(2x+(x^3 /6)))/(2x^4 )) =− lim_(x→0) ((x^4 (2+(x^2 /6)))/(12x^4 ))  = −(1/6) . (JS ♠)

Answered by mathmax by abdo last updated on 24/Jul/20

we have cosx =Σ_(n=0) ^∞  (((−1)^n x^(2n) )/((2n)!)) =1−(x^2 /2) +(x^4 /(4!)) +...  sinx =Σ_(n=0) ^∞  (((−1)^n  x^(2n+1) )/((2n+1)!)) =x−(x^3 /(3!)) +... ⇒  cos(sinx) =cos(x−(x^3 /(3!)) +...) =1−(1/2)(x−(x^3 /(3!)))^2  +(1/(4!))(x−(x^3 /(3!)))^4  +...  =1−(x^2 /2)(1−(x^2 /(3!)))^2  +(x^4 /(4!))(1−(x^2 /(3!)))^4  +....  =1−(x^2 /2)(1−2(x^2 /(3!))+(x^4 /((3!)^2 ))) +(x^4 /(4!))(1−(x^2 /(3!)))^2 (1−(x^2 /(3!)))^2 +...  =1−(x^2 /2) +(x^4 /(3!)) −(x^6 /(2(3!)^2 )) +(x^4 /(4!))(1−2(x^2 /(3!)) +(x^4 /((3!)^2 )))(1−2(x^2 /(3!)) +(x^4 /((3!)^2 )))  =1−(x^2 /2) +(x^4 /(3!))−(x^6 /(2(3!)^2 )) +(x^4 /(4!))(1−((2x^2 )/(3!)) +(x^4 /((3!)^2 )))−((2x^2 )/(3!)) +...)  ∼1−(x^2 /2) +((1/(3!))+(1/(4!)))x^4  ⇒cos(sinx)−cosx  ∼1−(x^2 /2)+((1/(3!))+(1/(4!)))x^4 −1+(x^2 /2)−(x^4 /(4!))  =(x^4 /(3!)) ⇒ lim_(x→0)   ((cos(sinx)−cosx)/x^4 ) =(1/(3!)) =(1/6)