Menu Close

lim-x-0-1-mx-n-1-nx-m-x-2-m-n-N-

Tinku Tara 0 Comments
Pin




Question Number 151979 by mathdanisur last updated on 24/Aug/21
lim_(x→0) (((1+mx)^n - (1+nx)^m )/x^2 ) = ? ; m;n∈N
limx→0(1+mx)n−(1+nx)mx2=?;m;n∈N
Answered by mr W last updated on 24/Aug/21
(1+mx)^n =1+nmx+((n(n−1)m^2 )/2)x^2 +o(x^2 ) (1+nx)^m =1+mnx+((m(m−1)n^2 )/2)x^2 +o(x^2 ) ⇒L=((n(n−1)m^2 )/2)−((m(m−1)n^2 )/2) ⇒L=((mn(n−m))/2)
(1+mx)n=1+nmx+n(n−1)m22x2+o(x2)(1+nx)m=1+mnx+m(m−1)n22x2+o(x2)⇒L=n(n−1)m22−m(m−1)n22⇒L=mn(n−m)2
Commented by mathdanisur last updated on 24/Aug/21
Thank You Ser
ThankYouSer
Commented by mathdanisur last updated on 25/Aug/21
Thank you Ser
ThankyouSer
Answered by mr W last updated on 24/Aug/21
L=((n(1+mx)^(n−1) m−m(1+nx)^(m−1) n)/(2x)) L=((n(n−1)(1+mx)^(n−2) m^2 −m(m−1)(1+nx)^(m−2) n^2 )/2) L=((n(n−1)m^2 −m(m−1)n^2 )/2) L=((mn(n−m))/2)
L=n(1+mx)n−1m−m(1+nx)m−1n2xL=n(n−1)(1+mx)n−2m2−m(m−1)(1+nx)m−2n22L=n(n−1)m2−m(m−1)n22L=mn(n−m)2
Answered by Kamel last updated on 25/Aug/21
With classic method: L=lim_(x→0) (((1+mx)^n −(1+nx)^m )/x^2 ), m,n∈N. =lim_(x→0) ((((1+mx)^n −1)−((1+nx)^m −1))/x^2 ) =lim_(x→0) ((mx(1+(1+mx)+(1+mx)^2 +...+(1+mx)^(n−1) )−nx(1+1+nx+(1+nx)^2 +...(1+nx)^(m−1) ))/x^2 ) =lim_(x→0) ((m(1+(1+mx)+(1+mx)^2 +...+(1+mx)^(n−1) )−n(1+1+nx+(1+nx)^2 +...(1+nx)^(m−1) ))/x) =lim_(x→0) ((m(1+(1+mx)+(1+mx)^2 +...+(1+mx)^(n−1) −n)−n(1+1+nx+(1+nx)^2 +...(1+nx)^(m−1) −m))/x) =lim_(x→0) [m(((1+mx−1)/x)+(((1+mx)^2 −1)/x)+...+(((1+mx)^(n−1) −1)/x))−n(((1+nx−1)/x)+(((1+nx)^2 −1)/x)+...+(((1+nx)^(m−1) −1)/x))] lim_(x→0) (((1+ax)^p −1)/x)=((ax(1+1+ax+(1+ax)^2 +...+(1+ax)^(p−1) )/x)=ap ∴ L=m(m+2m+3m+...+(n−1)m)−n(n+2n+3n+...n(m−1)) =m^2 ((n(n−1))/2)−n^2 ((m(m−1))/2) =((mn)/2)(nm−m−nm+n)=((mn(n−m))/2) ∴ lim_(x→0) (((1+mx)^n −(1+nx)^m )/x^2 )=((mn(n−m))/2) KAMEL BENAICHA
Withclassicmethod:L=limx→0(1+mx)n−(1+nx)mx2,m,n∈N.=limx→0((1+mx)n−1)−((1+nx)m−1)x2=limx→0mx(1+(1+mx)+(1+mx)2+…+(1+mx)n−1)−nx(1+1+nx+(1+nx)2+…(1+nx)m−1)x2=limx→0m(1+(1+mx)+(1+mx)2+…+(1+mx)n−1)−n(1+1+nx+(1+nx)2+…(1+nx)m−1)x=limx→0m(1+(1+mx)+(1+mx)2+…+(1+mx)n−1−n)−n(1+1+nx+(1+nx)2+…(1+nx)m−1−m)x=limx→0[m(1+mx−1x+(1+mx)2−1x+…+(1+mx)n−1−1x)−n(1+nx−1x+(1+nx)2−1x+…+(1+nx)m−1−1x)]limx→0(1+ax)p−1x=ax(1+1+ax+(1+ax)2+…+(1+ax)p−1x=ap∴L=m(m+2m+3m+…+(n−1)m)−n(n+2n+3n+…n(m−1))=m2n(n−1)2−n2m(m−1)2=mn2(nm−m−nm+n)=mn(n−m)2∴limx→0(1+mx)n−(1+nx)mx2=mn(n−m)2KAMELBENAICHA
Commented by mathdanisur last updated on 25/Aug/21
Thank you Ser
ThankyouSer

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *