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calculate-lim-x-0-x-ln-1-e-sinx-




Question Number 41761 by math khazana by abdo last updated on 12/Aug/18
calculate lim_(x→0)   x ln(1−e^(sinx) )
calculatelimx0xln(1esinx)
Answered by alex041103 last updated on 12/Aug/18
As x→0 the expression becomes one  of the form 0×∞ which is easilu  transformed in(∞/∞):  L=lim_(x→0)  ((ln(1−e^(sin x) ))/(1/x))=L′Hopital  =lim_(x→0) (((−e^(sinx) cosx)/(1−e^(sin x) ))/(−1/x^2 ))=lim_(x→0) (x^2 /(1−e^(sin(x)) ))cosxe^(sinx) =  =lim_(x→0) (x^2 /(1−e^(sinx) ))1  Now we apply L′Hopital until we find  a result:  L=lim_(x→0) ((2x)/(−e^(sin(x)) cos(x)))=((2×0)/(−e^(sin(0)) cos(0)))=  =(0/(1×1))=0  ⇒ lim_(x→0)   x ln(1−e^(sinx) )=0
Asx0theexpressionbecomesoneoftheform0×whichiseasilutransformedin:L=limx0ln(1esinx)1/x=LHopital=limx0esinxcosx1esinx1/x2=limx0x21esin(x)cosxesinx==limx0x21esinx1NowweapplyLHopitaluntilwefindaresult:L=limx02xesin(x)cos(x)=2×0esin(0)cos(0)==01×1=0limx0xln(1esinx)=0

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