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Question Number 197292 by universe last updated on 12/Sep/23

$$\underset{n\to \mathrm{\infty }}{\lim }{\int }_0^1\frac{{nx}^{n-1}}{1+x}dx=?$$

Answered by witcher3 last updated on 12/Sep/23

$$2\mathrm{x}⩽1+\mathrm{x}⩽2,\mathrm{\forall }\mathrm{x}\in [0,1]$$$$\Rightarrow \frac{{\mathrm{nx}}^{\mathrm{n}-1}}{2}⩽\frac{{\mathrm{nx}}^{\mathrm{n}-1}}{1+\mathrm{x}}⩽\frac{{\mathrm{nx}}^{\mathrm{n}-2}}{2}$$$$\frac{\mathrm{n}}{2\mathrm{n}}=\frac{1}{2}⩽{\int }_0^1\frac{{\mathrm{nx}}^{\mathrm{n}-1}}{1+\mathrm{x}}\mathrm{dx}⩽\frac{\mathrm{n}}{2(\mathrm{n}-1)}$$$$\Rightarrow \underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\int }_0^1{\mathrm{nx}}^{\mathrm{n}-1}{(1+\mathrm{x})}^{-1}\mathrm{dx}=\frac{1}{2}$$$$$$

Answered by witcher3 last updated on 12/Sep/23

$$\mathrm{Meth}2$$$$\mathrm{by}\mathrm{part}$$$${=\frac{{\mathrm{x}}^{\mathrm{n}}}{1+\mathrm{x}}]}_0^1+{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}}{{(1+\mathrm{x})}^2}\mathrm{dx}=\frac{1}{2}+{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}}{{(1+\mathrm{x})}^2}\mathrm{dx}$$$$\mathrm{by}\mathrm{part}2$$$${\int }_0^1{\mathrm{x}}^{\mathrm{n}}{(1+\mathrm{x})}^2\mathrm{dx}=\frac{1}{4(\mathrm{n}+1)}+\frac{2}{\mathrm{n}+1}{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}+1}}{{(1+\mathrm{x})}^3}\mathrm{dx}\to 0$$$$\mathrm{integral}\to \frac{1}{2}$$

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