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Question Number 151979 by mathdanisur last updated on 24/Aug/21

$$\underset{\mathbf{x}\to 0}{\lim }\frac{{(1+\mathrm{mx})}^{\mathbf{n}}-{(1+\mathrm{nx})}^{\mathbf{m}}}{{\mathrm{x}}^2}=?;\mathrm{m};\mathrm{n}\in \mathbb{N}$$

Answered by mr W last updated on 24/Aug/21

$${(1+mx)}^n=1+nmx+\frac{n(n-1)m^2}{2}{x}^2+o\left(x^2\right)$$$${(1+nx)}^m=1+mnx+\frac{m(m-1)n^2}{2}{x}^2+o\left(x^2\right)$$$$\Rightarrow L=\frac{n(n-1)m^2}{2}-\frac{m(m-1)n^2}{2}$$$$\Rightarrow L=\frac{mn(n-m)}{2}$$

Commented by mathdanisur last updated on 24/Aug/21

$$\mathrm{Thank}\mathrm{You}\mathbf{S}\mathrm{er}$$

Commented by mathdanisur last updated on 25/Aug/21

$$ThankyouSer$$

Answered by mr W last updated on 24/Aug/21

$$L=\frac{n{(1+mx)}^{n-1}m-m{(1+nx)}^{m-1}n}{2x}$$$$L=\frac{n(n-1){(1+mx)}^{n-2}{m}^2-m(m-1){(1+nx)}^{m-2}{n}^2}{2}$$$$L=\frac{n(n-1)m^2-m(m-1)n^2}{2}$$$$L=\frac{mn(n-m)}{2}$$

Answered by Kamel last updated on 25/Aug/21

$$Withclassicmethod:$$$$L=\underset{x\to 0}{lim}\frac{{(1+mx)}^n-{(1+nx)}^m}{x^2},m,n\in \mathbb{N}.$$$$=\underset{x\to 0}{lim}\frac{({(1+mx)}^n-1)-({(1+nx)}^m-1)}{x^2}$$$$=\underset{x\to 0}{lim}\frac{mx(1+(1+mx)+{(1+mx)}^2+...+{(1+mx)}^{n-1})-nx(1+1+nx+{(1+nx)}^2+...{(1+nx)}^{m-1})}{x^2}$$$$=\underset{x\to 0}{lim}\frac{m(1+(1+mx)+{(1+mx)}^2+...+{(1+mx)}^{n-1})-n(1+1+nx+{(1+nx)}^2+...{(1+nx)}^{m-1})}{x}$$$$=\underset{x\to 0}{lim}\frac{m(1+(1+mx)+{(1+mx)}^2+...+{(1+mx)}^{n-1}-n)-n(1+1+nx+{(1+nx)}^2+...{(1+nx)}^{m-1}-m)}{x}$$$$=\underset{x\to 0}{lim}[m(\frac{1+mx-1}{x}+\frac{{(1+mx)}^2-1}{x}+...+\frac{{(1+mx)}^{n-1}-1}{x})-n(\frac{1+nx-1}{x}+\frac{{(1+nx)}^2-1}{x}+...+\frac{{(1+nx)}^{m-1}-1}{x})]$$$$\underset{x\to 0}{lim}\frac{{(1+ax)}^p-1}{x}=\frac{ax(1+1+ax+{(1+ax)}^2+...+{(1+ax)}^{p-1}}{x}=ap$$$$\therefore L=m(m+2m+3m+...+(n-1)m)-n(n+2n+3n+...n(m-1))$$$$=m^2\frac{n(n-1)}{2}-n^2\frac{m(m-1)}{2}$$$$=\frac{mn}{2}(nm-m-nm+n)=\frac{mn(n-m)}{2}$$$$\therefore \underset{x\to 0}{\mathit{l}\mathit{i}\mathit{m}}\frac{{(1+\mathit{m}\mathit{x})}^{\mathit{n}}-{(1+\mathit{n}\mathit{x})}^{\mathit{m}}}{{\mathit{x}}^2}=\frac{\mathit{m}\mathit{n}(\mathit{n}-\mathit{m})}{2}$$$$\mathit{K}\mathit{A}\mathit{M}\mathit{E}\mathit{L}\mathit{B}\mathit{E}\mathit{N}\mathit{A}\mathit{I}\mathit{C}\mathit{H}\mathit{A}$$

Commented by mathdanisur last updated on 25/Aug/21

$$ThankyouSer$$

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