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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

$${Use}\:{L}'{Hopital}'{s}\:{Rule}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{mn}\left(\mathrm{1}+{nx}\right)^{{m}−\mathrm{1}} −{mn}\left(\mathrm{1}+{mx}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{1}+{mx}\right)^{\frac{\mathrm{1}}{{m}}−\mathrm{1}} −\left(\mathrm{1}+{nx}\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} } \\ $$$${again}\:{use}\:{L}'{Hopital}'{sRule} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{mn}\left({m}−\mathrm{1}\right){n}\left(\mathrm{1}+{nx}\right)^{{m}−\mathrm{2}} −{mn}\left({n}−\mathrm{1}\right){m}\left(\mathrm{1}+{mx}\right)^{{n}−\mathrm{2}} }{\left(\frac{\mathrm{1}}{{m}}−\mathrm{1}\right){m}\left(\mathrm{1}+{mx}\right)^{\frac{\mathrm{1}}{{m}}−\mathrm{2}} −\left(\frac{\mathrm{1}}{{n}}−\mathrm{1}\right){n}\left(\mathrm{1}+{nx}\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{2}} } \\ $$$$\frac{{mn}\left({m}−\mathrm{1}\right)−{mn}\left({n}−\mathrm{1}\right)}{\left(\frac{\mathrm{1}}{{m}}−\mathrm{1}\right){m}−\left(\frac{\mathrm{1}}{{n}}−\mathrm{1}\right){n}} \\ $$$$=−{mn} \\ $$

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

$${method}\left({II}\right) \\ $$$$\left(\mathrm{1}+{nx}\right)^{{m}} =\mathrm{1}+{mnx}+{o}\left({x}^{\mathrm{2}} \right) \\ $$$$\left(\mathrm{1}+{mx}\right)^{{n}} =\mathrm{1}+{nmx}+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${So}\: \\ $$$$\left(\mathrm{1}+{nx}\right)^{{m}} −\left(\mathrm{1}+{mx}\right)^{{n}} =\mathrm{0} \\ $$$${for}\:{second}\:{order}\:{term} \\ $$$$\left(\mathrm{1}+{nx}\right)^{{n}} =\mathrm{1}+{mnx}+\frac{{m}\left({m}−\mathrm{1}\right)}{\mathrm{2}}\left({nx}\right)^{\mathrm{2}} +{o}\left({x}^{\mathrm{3}} \right) \\ $$$$\left(\mathrm{1}+{mx}\right)^{{n}} =\mathrm{1}+{mnx}\:+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\left({mx}\right)^{\mathrm{2}} +{o}\left({x}^{\mathrm{3}} \right) \\ $$$$\left(\mathrm{1}+{nx}\right)^{{n}} −\left(\mathrm{1}+{mx}\right)^{{m}} =\frac{{m}\left({m}−\mathrm{1}\right)}{\mathrm{2}}\left({nx}\right)^{\mathrm{2}} −\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\left({mx}\right)^{\mathrm{2}} \\ $$$${similarly}\:{do}\:{for}\:{denominator} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{m}\left({m}−\mathrm{1}\right)}{\mathrm{2}}\left({nx}\right)^{\mathrm{2}} −\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\left({mx}\right)^{\mathrm{2}} }{\frac{\left({m}−\mathrm{1}\right)}{\mathrm{2}{m}}\left({mx}\right)^{\mathrm{2}} −\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}{n}}\left({nx}\right)^{\mathrm{2}} } \\ $$$$=−{mn} \\ $$

Commented by Spillover last updated on 24/Sep/24

thanks

$${thanks} \\ $$

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