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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+x^(2018) )=x^(2018) −(1/2)x^(4036) +... (ln(1+x))^(2018) =(x−(1/2)x^2 +...)^(2018) =x^(2018) + (((2018)),(( 1)) )x^(2017) (−(1/2)x^2 )+... =x^(2018) −1009x^(2019) +... ⇒lim_(x→0) ((ln(1+x^(2018) )−(ln(1+x))^(2018) )/x^(2019) )=lim_(x→0) ((1009x^(2019) +o(x^(2019) ))/x^(2019) )=1009

$$\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2018}} \right)=\mathrm{x}^{\mathrm{2018}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{4036}} +... \\ $$$$\left(\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\right)^{\mathrm{2018}} =\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +...\right)^{\mathrm{2018}} =\mathrm{x}^{\mathrm{2018}} +\begin{pmatrix}{\mathrm{2018}}\\{\:\:\:\:\mathrm{1}}\end{pmatrix}\mathrm{x}^{\mathrm{2017}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right)+... \\ $$$$=\mathrm{x}^{\mathrm{2018}} −\mathrm{1009x}^{\mathrm{2019}} +... \\ $$$$\Rightarrow\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2018}} \right)−\left(\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\right)^{\mathrm{2018}} }{\mathrm{x}^{\mathrm{2019}} }=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1009x}^{\mathrm{2019}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{2019}} \right)}{\mathrm{x}^{\mathrm{2019}} }=\mathrm{1009} \\ $$

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