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Question Number 40709 by math khazana by abdo last updated on 26/Jul/18
calculate lim_(x→0)    ((cos(x−sinx)−1)/(x^2  ))
calculatelimx0cos(xsinx)1x2
Commented by maxmathsup by imad last updated on 26/Jul/18
  we have sinx =Σ_(n=0) ^∞  (((−1)^n )/((2n+1)!))x^(2n+1)  ⇒sinx =x−(1/(3!)) x^3  +o(x^5 ) ⇒  x−sinx =(x^3 /6) +o(x^5 ) ⇒cos(x−sinx) =cos((x^3 /6) +o(x^5 ))∼1−(x^6 /(12)) (x→0) ⇒  cos(x−sinx)−1 ∼−(x^6 /(12)) ⇒ ((cos(x−sinx)−1)/x^2 ) ∼ (x^4 /(12)) ⇒  lim_(x→0)     ((cos(x−sinx)−1)/x^2 ) =0
wehavesinx=n=0(1)n(2n+1)!x2n+1sinx=x13!x3+o(x5)xsinx=x36+o(x5)cos(xsinx)=cos(x36+o(x5))1x612(x0)cos(xsinx)1x612cos(xsinx)1x2x412limx0cos(xsinx)1x2=0
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18
lim_(x→0)  ((−2sin^2 (((x−sinx)/2)))/((((x−sinx)/2))^2 ))×(((((x−sinx)/2))^2 )/x^2 )  when x→0   ((x−sinx)/2)→0  t=((x−sinx)/2)  lim_(t→0)  ((−2sin^2 t)/t^2 )×  lim_(x→0) ((x^2 (1−((sinx)/x))^2 )/(4x^2 ))  =−2×(1/4)×(1−1)=0
limx02sin2(xsinx2)(xsinx2)2×(xsinx2)2x2whenx0xsinx20t=xsinx2limt02sin2tt2×limx0x2(1sinxx)24x2=2×14×(11)=0

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