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Question Number 211917 by Spillover last updated on 24/Sep/24

Answered by BHOOPENDRA last updated on 24/Sep/24

Use L′Hopital′s Rule lim_(x→0) ((mn(1+nx)^(m−1) −mn(1+mx)^(n−1) )/((1+mx)^((1/m)−1) −(1+nx)^((1/n)−1) )) again use L′Hopital′sRule lim_(x→0) ((mn(m−1)n(1+nx)^(m−2) −mn(n−1)m(1+mx)^(n−2) )/(((1/m)−1)m(1+mx)^((1/m)−2) −((1/n)−1)n(1+nx)^((1/n)−2) )) ((mn(m−1)−mn(n−1))/(((1/m)−1)m−((1/n)−1)n)) =−mn

UseL′Hopital′sRulelimx→0mn(1+nx)m−1−mn(1+mx)n−1(1+mx)1m−1−(1+nx)1n−1againuseL′Hopital′sRulelimx→0mn(m−1)n(1+nx)m−2−mn(n−1)m(1+mx)n−2(1m−1)m(1+mx)1m−2−(1n−1)n(1+nx)1n−2mn(m−1)−mn(n−1)(1m−1)m−(1n−1)n=−mn

Commented by Spillover last updated on 24/Sep/24

great.thanks

great.thanks

Answered by BHOOPENDRA last updated on 24/Sep/24

method(II) (1+nx)^m =1+mnx+o(x^2 ) (1+mx)^n =1+nmx+o(x^2 ) So (1+nx)^m −(1+mx)^n =0 for second order term (1+nx)^n =1+mnx+((m(m−1))/2)(nx)^2 +o(x^3 ) (1+mx)^n =1+mnx +((n(n−1))/2)(mx)^2 +o(x^3 ) (1+nx)^n −(1+mx)^m =((m(m−1))/2)(nx)^2 −((n(n−1))/2)(mx)^2 similarly do for denominator lim_(x→0) ((((m(m−1))/2)(nx)^2 −((n(n−1))/2)(mx)^2 )/((((m−1))/(2m))(mx)^2 −((n(n−1))/(2n))(nx)^2 )) =−mn

method(II)(1+nx)m=1+mnx+o(x2)(1+mx)n=1+nmx+o(x2)So(1+nx)m−(1+mx)n=0forsecondorderterm(1+nx)n=1+mnx+m(m−1)2(nx)2+o(x3)(1+mx)n=1+mnx+n(n−1)2(mx)2+o(x3)(1+nx)n−(1+mx)m=m(m−1)2(nx)2−n(n−1)2(mx)2similarlydofordenominatorlimx→0m(m−1)2(nx)2−n(n−1)2(mx)2(m−1)2m(mx)2−n(n−1)2n(nx)2=−mn

Commented by Spillover last updated on 24/Sep/24

thanks

thanks

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