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Question Number 172910 by DAVONG last updated on 03/Jul/22
lim_(x→1) ((e−x^(1/(x−1)) )/(x−1))=?
limx→1e−x1x−1x−1=?
Answered by FongXD last updated on 03/Jul/22
L=lim_(x→1) ((e−e^((1/(x−1))lnx) )/(x−1))=−lim_(x→1) ((e(e^(((lnx)/(x−1))−1) −1))/(((lnx)/(x−1))−1))×((((lnx)/(x−1))−1)/(x−1)) L=−elim_(x→1) ((lnx−x+1)/((x−1)^2 )) put x=e^t , if x→1, ⇒ t→0 then L=−elim_(t→0) ((t−e^t +1)/((e^t −1)^2 ))=elim_(t→0) ((e^t −t−1)/t^2 )×((t/(e^t −1)))^2 L=elim_(t→0) ((e^t −t−1)/t^2 )=(e/4)lim_(u→0) ((e^(2u) −2u−1)/u^2 ), [t=2u] L−(e/4)=(e/4)lim_(u→0) ((e^(2u) −u^2 −2u−1)/u^2 )=(e/4)lim_(u→0) ((e^(2u) −(u+1)^2 )/u^2 ) L−(e/4)=(e/4)lim_(u→0) ((e^u −u−1)/u^2 )(e^u +u+1) L−(e/4)=(1/4)L×2=(1/2)L L=(e/2)
L=limx→1e−e1x−1lnxx−1=−limx→1e(elnxx−1−1−1)lnxx−1−1×lnxx−1−1x−1L=−elimx→1lnx−x+1(x−1)2putx=et,ifx→1,⇒t→0thenL=−elimt→0t−et+1(et−1)2=elimt→0et−t−1t2×(tet−1)2L=elimt→0et−t−1t2=e4limu→0e2u−2u−1u2,[t=2u]L−e4=e4limu→0e2u−u2−2u−1u2=e4limu→0e2u−(u+1)2u2L−e4=e4limu→0eu−u−1u2(eu+u+1)L−e4=14L×2=12LL=e2

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