Question Number 151979 by mathdanisur last updated on 24/Aug/21
$$\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\mathrm{mx}\right)^{\boldsymbol{\mathrm{n}}} \:-\:\left(\mathrm{1}+\mathrm{nx}\right)^{\boldsymbol{\mathrm{m}}} }{\mathrm{x}^{\mathrm{2}} }\:=\:?\:\:;\:\:\mathrm{m};\mathrm{n}\in\mathbb{N} \\ $$
Answered by mr W last updated on 24/Aug/21
$$\left(\mathrm{1}+{mx}\right)^{{n}} =\mathrm{1}+{nmx}+\frac{{n}\left({n}−\mathrm{1}\right){m}^{\mathrm{2}} }{\mathrm{2}}{x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right) \\ $$$$\left(\mathrm{1}+{nx}\right)^{{m}} =\mathrm{1}+{mnx}+\frac{{m}\left({m}−\mathrm{1}\right){n}^{\mathrm{2}} }{\mathrm{2}}{x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{L}=\frac{{n}\left({n}−\mathrm{1}\right){m}^{\mathrm{2}} }{\mathrm{2}}−\frac{{m}\left({m}−\mathrm{1}\right){n}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{L}=\frac{{mn}\left({n}−{m}\right)}{\mathrm{2}} \\ $$
Commented by mathdanisur last updated on 24/Aug/21
$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$
Commented by mathdanisur last updated on 25/Aug/21
$${Thank}\:{you}\:{Ser} \\ $$
Answered by mr W last updated on 24/Aug/21
$${L}=\frac{{n}\left(\mathrm{1}+{mx}\right)^{{n}−\mathrm{1}} {m}−{m}\left(\mathrm{1}+{nx}\right)^{{m}−\mathrm{1}} {n}}{\mathrm{2}{x}} \\ $$$${L}=\frac{{n}\left({n}−\mathrm{1}\right)\left(\mathrm{1}+{mx}\right)^{{n}−\mathrm{2}} {m}^{\mathrm{2}} −{m}\left({m}−\mathrm{1}\right)\left(\mathrm{1}+{nx}\right)^{{m}−\mathrm{2}} {n}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${L}=\frac{{n}\left({n}−\mathrm{1}\right){m}^{\mathrm{2}} −{m}\left({m}−\mathrm{1}\right){n}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${L}=\frac{{mn}\left({n}−{m}\right)}{\mathrm{2}} \\ $$
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](https://www.tinkutara.com/question/Q152004.png)
$${With}\:{classic}\:{method}: \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{mx}\right)^{{n}} −\left(\mathrm{1}+{nx}\right)^{{m}} }{{x}^{\mathrm{2}} },\:{m},{n}\in\mathbb{N}. \\ $$$$\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\left(\mathrm{1}+{mx}\right)^{{n}} −\mathrm{1}\right)−\left(\left(\mathrm{1}+{nx}\right)^{{m}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} } \\ $$$$\:\:=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{mx}\left(\mathrm{1}+\left(\mathrm{1}+{mx}\right)+\left(\mathrm{1}+{mx}\right)^{\mathrm{2}} +…+\left(\mathrm{1}+{mx}\right)^{{n}−\mathrm{1}} \right)−{nx}\left(\mathrm{1}+\mathrm{1}+{nx}+\left(\mathrm{1}+{nx}\right)^{\mathrm{2}} +…\left(\mathrm{1}+{nx}\right)^{{m}−\mathrm{1}} \right)}{{x}^{\mathrm{2}} } \\ $$$$\:\:=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{m}\left(\mathrm{1}+\left(\mathrm{1}+{mx}\right)+\left(\mathrm{1}+{mx}\right)^{\mathrm{2}} +…+\left(\mathrm{1}+{mx}\right)^{{n}−\mathrm{1}} \right)−{n}\left(\mathrm{1}+\mathrm{1}+{nx}+\left(\mathrm{1}+{nx}\right)^{\mathrm{2}} +…\left(\mathrm{1}+{nx}\right)^{{m}−\mathrm{1}} \right)}{{x}} \\ $$$$\:\:=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{m}\left(\mathrm{1}+\left(\mathrm{1}+{mx}\right)+\left(\mathrm{1}+{mx}\right)^{\mathrm{2}} +…+\left(\mathrm{1}+{mx}\right)^{{n}−\mathrm{1}} −{n}\right)−{n}\left(\mathrm{1}+\mathrm{1}+{nx}+\left(\mathrm{1}+{nx}\right)^{\mathrm{2}} +…\left(\mathrm{1}+{nx}\right)^{{m}−\mathrm{1}} −{m}\right)}{{x}} \\ $$$$\:\:=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left[{m}\left(\frac{\mathrm{1}+{mx}−\mathrm{1}}{{x}}+\frac{\left(\mathrm{1}+{mx}\right)^{\mathrm{2}} −\mathrm{1}}{{x}}+…+\frac{\left(\mathrm{1}+{mx}\right)^{{n}−\mathrm{1}} −\mathrm{1}}{{x}}\right)−{n}\left(\frac{\mathrm{1}+{nx}−\mathrm{1}}{{x}}+\frac{\left(\mathrm{1}+{nx}\right)^{\mathrm{2}} −\mathrm{1}}{{x}}+…+\frac{\left(\mathrm{1}+{nx}\right)^{{m}−\mathrm{1}} −\mathrm{1}}{{x}}\right)\right] \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{ax}\right)^{{p}} −\mathrm{1}}{{x}}=\frac{{ax}\left(\mathrm{1}+\mathrm{1}+{ax}+\left(\mathrm{1}+{ax}\right)^{\mathrm{2}} +…+\left(\mathrm{1}+{ax}\right)^{{p}−\mathrm{1}} \right.}{{x}}={ap} \\ $$$$\therefore\:{L}={m}\left({m}+\mathrm{2}{m}+\mathrm{3}{m}+…+\left({n}−\mathrm{1}\right){m}\right)−{n}\left({n}+\mathrm{2}{n}+\mathrm{3}{n}+…{n}\left({m}−\mathrm{1}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:={m}^{\mathrm{2}} \frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}−{n}^{\mathrm{2}} \frac{{m}\left({m}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{{mn}}{\mathrm{2}}\left({nm}−{m}−{nm}+{n}\right)=\frac{{mn}\left({n}−{m}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\therefore\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\frac{\left(\mathrm{1}+\boldsymbol{{mx}}\right)^{\boldsymbol{{n}}} −\left(\mathrm{1}+\boldsymbol{{nx}}\right)^{\boldsymbol{{m}}} }{\boldsymbol{{x}}^{\mathrm{2}} }=\frac{\boldsymbol{{mn}}\left(\boldsymbol{{n}}−\boldsymbol{{m}}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{KAMEL}}\:\boldsymbol{{BENAICHA}} \\ $$
Commented by mathdanisur last updated on 25/Aug/21
$${Thank}\:{you}\:{Ser} \\ $$