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Question Number 168555 by mathlove last updated on 13/Apr/22

Answered by LEKOUMA last updated on 13/Apr/22

$$\underset{x\to 0}{\lim }\frac{-x^2-\frac{x^4}{3}+x^2}{x^4}=\underset{x\to 0}{\lim }\frac{-x^4}{3x^4}=-\frac{1}{3}$$

Answered by qaz last updated on 13/Apr/22

$$\ln (1-\mathrm{x})\ln (1+\mathrm{x})+{\mathrm{x}}^2=-(\mathrm{x}+\frac{1}{2}{\mathrm{x}}^2+\frac{1}{3}{\mathrm{x}}^3+...)(\mathrm{x}-\frac{1}{2}{\mathrm{x}}^2+\frac{1}{3}{\mathrm{x}}^3+...)+{\mathrm{x}}^2$$$$=-[{(\mathrm{x}+\frac{1}{3}{\mathrm{x}}^3)}^2-\frac{1}{4}{\mathrm{x}}^4+...]+{\mathrm{x}}^2=-({\mathrm{x}}^2+\frac{2}{3}{\mathrm{x}}^4-\frac{1}{4}{\mathrm{x}}^4+...)+{\mathrm{x}}^2=-\frac{5}{12}{\mathrm{x}}^4+\mathrm{o}\left({\mathrm{x}}^4\right)$$$$\Rightarrow \underset{\mathrm{x}\to 0}{\lim }\frac{\ln (1-\mathrm{x})\ln (1+\mathrm{x})+{\mathrm{x}}^2}{{\mathrm{x}}^4}=\underset{\mathrm{x}-0}{\lim }\frac{-\frac{5}{12}{\mathrm{x}}^4+\mathrm{o}\left({\mathrm{x}}^4\right)}{{\mathrm{x}}^4}=-\frac{5}{12}$$

Commented by mathlove last updated on 17/Apr/22

$$thanks$$

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