help with solving a limit with logarithms

For all $-1<x<1$ we have $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+...,$$ which says $$\frac{\ln\left(1+x^{2018}\right)-\ln^{2018}(1+x)}{x^{2019}}=\frac{x^{2018}-\frac{x^{4036}}{2}+...-\left(x-\frac{x^2}{2}+...\right)^{2018}}{x^{2019}}=$$ $$=\frac{1-\frac{x^{2018}}{2}+...-\left(1-\frac{x}{2}+...\right)^{2018}}{x}=\frac{1009x-\frac{2018\cdot2017x^2}{8}+...}{x}\rightarrow1009.$$