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Question Number 156788 by mathlove last updated on 15/Oct/21

$$\underset{x\to 1}{\lim }\frac{x+x^2+x^3+.....x^n-n}{x+x^2+x^3+.....x^m-m}=?\frac{0}{0}$$

Answered by puissant last updated on 15/Oct/21

$$\mathcal{L}=\underset{x\to 1}{\lim }\frac{x+x^2+x^3+....+x^n-n}{x+x^2+x^3+....+x^m-m}=\frac{0}{0}$$$$hospitalrule\to$$$$\mathcal{L}=\underset{x\to 1}{\lim }\frac{1x^0+2x+3x^2+....+{nx}^{n-1}}{1x^0+2x+3x^3+.....+{mx}^{m-1}}$$$$\Rightarrow \mathcal{L}=\frac{1+2+....+n}{1+2+....+m}=\frac{\frac{n(n+1)}{2}}{\frac{m(m+1)}{2}}$$$$\therefore \because \mathcal{L}=\frac{n(n+1)}{m(m+1)}..$$

Commented by mathlove last updated on 15/Oct/21

$$weathouthospitalrulseplissir$$

Commented by puissant last updated on 15/Oct/21

$$Thehospitalruleisusedwhenthe$$$$limittendstofinitenumberand$$$$whenwehave0/0ifnotwecannot$$$$useit...$$

Answered by FongXD last updated on 15/Oct/21

$$\mathrm{Without}{\mathrm{L}}^{\prime }{\mathrm{hospital}}^{\prime }\mathrm{s}\mathrm{rule}$$$$=\underset{\mathrm{x}\to 1}{\lim }\frac{(\mathrm{x}-1)+({\mathrm{x}}^2-1)+({\mathrm{x}}^3-1)+...+({\mathrm{x}}^{\mathrm{n}}-1)}{(\mathrm{x}-1)+({\mathrm{x}}^2-1)+({\mathrm{x}}^3-1)+...+({\mathrm{x}}^{\mathrm{m}}-1)}$$$$=\underset{\mathrm{x}\to 1}{\lim }\frac{(\mathrm{x}-1)[1+(\mathrm{x}+1)+({\mathrm{x}}^2+\mathrm{x}+1)+...+({\mathrm{x}}^{\mathrm{n}-1}+{\mathrm{x}}^{\mathrm{n}-2}+...+\mathrm{x}+1)]}{(\mathrm{x}-1)[1+(\mathrm{x}+1)+({\mathrm{x}}^2+\mathrm{x}+1)+...+({\mathrm{x}}^{\mathrm{m}-1}+{\mathrm{x}}^{\mathrm{m}-2}+...+\mathrm{x}+1)]}$$$$=\underset{\mathrm{x}\to 1}{\lim }\frac{1+(\mathrm{x}+1)+({\mathrm{x}}^2+\mathrm{x}+1)+...+({\mathrm{x}}^{\mathrm{n}-1}+{\mathrm{x}}^{\mathrm{n}-2}+...+\mathrm{x}+1)}{1+(\mathrm{x}+1)+({\mathrm{x}}^2+\mathrm{x}+1)+...+({\mathrm{x}}^{\mathrm{m}-1}+{\mathrm{x}}^{\mathrm{m}-2}+...+\mathrm{x}+1)}$$$$=\frac{1+2+3+...+\mathrm{n}}{1+2+3+...+\mathrm{m}}=\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2}}{\frac{\mathrm{m}(\mathrm{m}+1)}{2}}=\frac{\mathrm{n}(\mathrm{n}+1)}{\mathrm{m}(\mathrm{m}+1)}$$

Commented by mathlove last updated on 16/Oct/21

$$thankssir$$

Commented by Rasheed.Sindhi last updated on 16/Oct/21

$$\mathcal{N}ice\mathcal{S}ir!$$

Answered by mindispower last updated on 16/Oct/21

$$x+x^2+x^3+.....+x^n-n=$$$$(x-1)(x^{n-1}+2x^{n-2}+..{kx}^{k-1}+....+n)$$$$x+x^2+.....+x^m-m=(x-1)(x^{m-1}+.2x^{m-2}+.....+m)$$$$\iff \underset{x\to 0}{\lim }.\frac{\mathit{n}+(\mathit{n}-1)x+...(k-1)x^{k-1}+...+\mathit{x}}{\mathit{m}+(\mathit{m}-1)+....+(\mathit{k}-1){\mathit{x}}^{\mathit{k}-1}+..+\mathit{x}}$$$$=\frac{\underset{k=1}{\sum ^n}k}{\underset{k=1}{\sum ^m}k}=\frac{n(n+1)}{m(m+1)}$$

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