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calculate-lim-x-0-ln-1-x-sinx-ln-1-sin-2x-x-2-




Question Number 62200 by maxmathsup by imad last updated on 17/Jun/19
calculate lim_(x→0)   ((ln(1+x+sinx)−ln(1+sin(2x)))/x^2 )
calculatelimx0ln(1+x+sinx)ln(1+sin(2x))x2
Commented by maxmathsup by imad last updated on 18/Jun/19
let use hospital theorem let u(x) =ln(1+x+sinx)−ln(1+sin(2x) ⇒  u^′ (x) =((1+cosx)/(1+x+sinx)) −((2cosx)/(1+sin(2x)))  u^(′′) (x) =((−sinx(1+x +sinx)−(1+cosx)^2 )/((1+x+sinx)^2 )) −2 ((−sinx(1+sin(2x))−cosx(2cosx(2x)))/((1+sin(2x))^2 ))  ⇒lim_(x→0)    u^((2)) (x) =((−4)/1) −2 ((−2)/1) =−4+4 =0  v(x)=x^2  ⇒v^′ (x) =2x ⇒v^((2)) (x)=2  ⇒lim_(x→0)    ((ln(1+x+sinx)−ln(1+sin(2x)))/x^2 )  =lim_(x→0)    ((u^((2)) (x))/(v^((2)) (x))) =0
letusehospitaltheoremletu(x)=ln(1+x+sinx)ln(1+sin(2x)u(x)=1+cosx1+x+sinx2cosx1+sin(2x)u(x)=sinx(1+x+sinx)(1+cosx)2(1+x+sinx)22sinx(1+sin(2x))cosx(2cosx(2x))(1+sin(2x))2limx0u(2)(x)=41221=4+4=0v(x)=x2v(x)=2xv(2)(x)=2limx0ln(1+x+sinx)ln(1+sin(2x))x2=limx0u(2)(x)v(2)(x)=0
Answered by tanmay last updated on 17/Jun/19
lim_(x→0)  ((ln(((1+x+sinx)/(1+sin2x))))/x^2 )  lim_(x→0)  ((ln(1+((1+x+sinx)/(1+sin2x))−1))/((((1+x+sinx)/(1+sin2x))−1)))×(((x+sinx−sin2x)/(1+sin2x))/x^2 )  lim_(t→0)  ((ln(1+t))/t)×lim_(x→0) (1/(1+sin2x))×lim_(x→0)  ((x+sinx−sin2x)/x^2 )  1×1×lim_(x→0)  ((1+cosx−2cos2x)/(2x))((0/0))  =lim_(x→0)  ((−sinx+4sin2x)/2)=0
limx0ln(1+x+sinx1+sin2x)x2limx0ln(1+1+x+sinx1+sin2x1)(1+x+sinx1+sin2x1)×x+sinxsin2x1+sin2xx2limt0ln(1+t)t×limx011+sin2x×limx0x+sinxsin2xx21×1×limx01+cosx2cos2x2x(00)=limx0sinx+4sin2x2=0

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