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Question Number 132328 by liberty last updated on 13/Feb/21

$$\underset{x\to 1}{\lim }\frac{\mathrm{p}-\mathrm{x}-{\mathrm{x}}^2-{\mathrm{x}}^3-...-{\mathrm{x}}^{\mathrm{p}}}{\mathrm{x}-1}=?$$

Answered by EDWIN88 last updated on 13/Feb/21

$$\mathrm{The}\mathrm{limit}\mathrm{is}\mathrm{form}\frac{0}{0}$$$${\mathrm{L}}^{\prime }\mathrm{H}\widehat{\mathrm{o}pital}\Rightarrow \mathrm{L}=\underset{x\to 1}{\lim }\frac{-1-2\mathrm{x}-{3\mathrm{x}}^2-{4\mathrm{x}}^3-...-{\mathrm{px}}^{\mathrm{p}-1}}{1}$$$$\mathrm{L}=-1-2-3-4-...-\mathrm{p}$$$$\mathrm{L}=-\underset{\mathrm{AP}}{\underbrace{(1+2+3+4+...+\mathrm{p})}}=-\frac{\mathrm{p}(\mathrm{p}+1)}{2}$$

Answered by mathmax by abdo last updated on 13/Feb/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{p}-(\mathrm{x}+{\mathrm{x}}^2+....+{\mathrm{x}}^{\mathrm{p}})}{\mathrm{x}-1}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{cha}7\mathrm{gement}\mathrm{x}-1=\mathrm{t}(\mathrm{sot}\to \mathrm{o})$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(1+\mathrm{t})=\frac{\mathrm{p}-(1+\mathrm{t}+{(1+\mathrm{t})}^2+....{(1+\mathrm{t})}^{\mathrm{p}})}{\mathrm{t}}\Rightarrow$$$$\mathrm{f}(1+\mathrm{t})\sim \frac{\mathrm{p}-(1+\mathrm{t}+1+2\mathrm{t}+....+1+\mathrm{pt})}{\mathrm{t}}=\frac{\mathrm{p}-\mathrm{p}-(1+2+3...+\mathrm{p})\mathrm{t}}{\mathrm{t}}$$$$=-\frac{\mathrm{p}(\mathrm{p}+1)}{2}\Rightarrow {\lim }_{\mathrm{t}\to 0}\mathrm{f}(1+\mathrm{t})=-\frac{\mathrm{p}(\mathrm{p}+1)}{2}={\lim }_{\mathrm{x}\to 1}\mathrm{f}\left(\mathrm{x}\right)$$

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