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Question Number 213821 by hardmath last updated on 17/Nov/24

Find:   lim_(x→0)  (((sinx)/x))^((sinx)/(x − sinx))   =  ?

Find:limx0(sinxx)sinxxsinx=?

Answered by mehdee7396 last updated on 17/Nov/24

lim_(x→0) (((sinx)/x)−1)((sinx)/(x−sinx))  =lim_(x→0) (((sinx−x)/x))((sinx)/(x−sinx))=−1  ⇒answer=e^(−1)

limx0(sinxx1)sinxxsinx=limx0(sinxxx)sinxxsinx=1answer=e1

Answered by lepuissantcedricjunior last updated on 20/Nov/24

lim_(x→0) (((sinx)/x))^((sinx)/(x−sinx)) =1^(FI)   =>lim_(x→0) (((sinx)/x))^((sinx)/(x−sinx)) =lim_(x→0) e^((sinx ln(((sinx)/x)))/(x−sinx))   dl sinx a^�  l′ordre 3  au vois de 0  sinx=Σ_(k=0) ^n (x^k /(k!))×f^k (0)=x−(x^3 /6)+x(0)^3   lim_(x→0) e^(((x−(x^3 /6))ln(1−(x^2 /6)))/(x−x+(x^3 /6))) or dl ln(1−t) a^�  l′ordre 3 au vois de 0 donne  ln(1−t)=−t+(t^2 /2)−(t^3 /3)+t(0)^3   =>lim_(x→0) (((sinx)/x))^((sinx)/(x−sinx)) =lim_(x→0) e^(((x−(x^3 /6))(−(x^2 /6)+(x^4 /(12))))/(x^3 /6))   =lim_(x→0) e^((−(x^3 /6))/(x^3 /6)) =e^(−1)   ..........................................................  prof cedric junior.....

limx0(sinxx)sinxxsinx=1FI=>limx0(sinxx)sinxxsinx=limx0esinxln(sinxx)xsinxdlsinxa`lordre3auvoisde0sinx=nk=0xkk!×fk(0)=xx36+x(0)3limx0e(xx36)ln(1x26)xx+x36ordlln(1t)a`lordre3auvoisde0donneln(1t)=t+t22t33+t(0)3=>limx0(sinxx)sinxxsinx=limx0e(xx36)(x26+x412)x36=limx0ex36x36=e1..........................................................profcedricjunior.....

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