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Question Number 92167 by john santu last updated on 05/May/20

$$\underset{x\to \mathrm{\infty }}{\lim }\mathrm{x}-{\mathrm{x}}^2\ln (1+\frac{1}{\mathrm{x}})?$$

Commented by mathmax by abdo last updated on 06/May/20

$$wehave{ln}^{{}^{\prime }}(1+u)=\frac{1}{1+u}=1-u+o\left(u\right)\Rightarrow ln(1+u)=u-\frac{u^2}{2}+o\left(u^3\right)$$$$\Rightarrow ln(1+\frac{1}{x})=\frac{1}{x}-\frac{1}{2x^2}+o\left(\frac{1}{x^3}\right)(x\sim \mathrm{\infty })\Rightarrow$$$$x^{2}ln(1+\frac{1}{x})=x-\frac{1}{2}+o\left(\frac{1}{x}\right)\Rightarrow x-x^{2}ln(1+\frac{1}{x})$$$$=x-(x-\frac{1}{2}+o\left(\frac{1}{x}\right))=\frac{1}{2}+o\left(\frac{1}{x}\right)\Rightarrow {lim}_{x\to \mathrm{\infty }}x-x^{2}ln(1+\frac{1}{x})=\frac{1}{2}$$

Answered by jagoll last updated on 05/May/20

$$\ln (1+\frac{1}{\mathrm{x}})=\frac{1}{\mathrm{x}}-\frac{1}{{2\mathrm{x}}^2}+\frac{1}{{3\mathrm{x}}^3}-\frac{1}{{4\mathrm{x}}^4}+...$$$$\mid \frac{1}{\mathrm{x}}\mid <1$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim x}-{\mathrm{x}}^2(\frac{1}{\mathrm{x}}-\frac{1}{{2\mathrm{x}}^2}+\frac{1}{{3\mathrm{x}}^3}-\frac{1}{{4\mathrm{x}}^4}+...)=$$$$\underset{x\to \mathrm{\infty }}{\lim }\mathrm{x}-(\mathrm{x}-\frac{1}{2}+\frac{1}{3\mathrm{x}}-\frac{1}{{4\mathrm{x}}^2}+...)=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{2}-\frac{1}{3\mathrm{x}}+\frac{1}{{4\mathrm{x}}^2}-...=\frac{1}{2}$$

Commented by john santu last updated on 05/May/20

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