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Question Number 170197 by mathlove last updated on 18/May/22

Answered by qaz last updated on 18/May/22

$$\ln (1+{\mathrm{x}}^{2018})={\mathrm{x}}^{2018}-\frac{1}{2}{\mathrm{x}}^{4036}+...$$$${\left(\ln (1+\mathrm{x})\right)}^{2018}={(\mathrm{x}-\frac{1}{2}{\mathrm{x}}^2+...)}^{2018}={\mathrm{x}}^{2018}+\left(\begin{array}{c}2018\\ 1\end{array}\right){\mathrm{x}}^{2017}(-\frac{1}{2}{\mathrm{x}}^2)+...$$$$={\mathrm{x}}^{2018}-{1009\mathrm{x}}^{2019}+...$$$$\Rightarrow \underset{\mathrm{x}\to 0}{\lim }\frac{\ln (1+{\mathrm{x}}^{2018})-{\left(\ln (1+\mathrm{x})\right)}^{2018}}{{\mathrm{x}}^{2019}}=\underset{\mathrm{x}\to 0}{\lim }\frac{{1009\mathrm{x}}^{2019}+\mathrm{o}\left({\mathrm{x}}^{2019}\right)}{{\mathrm{x}}^{2019}}=1009$$

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