[tex]\lim\limits_{x\to 0}\dfrac{\ln(1+x^{2018})-\ln^{2018}(1+x)}{x^{2019}} = \\ \\ = \lim\limits_{x\to 0}\dfrac{\ln(1+x^{2018})}{x^{2019}}- \lim\limits_{x\to 0}\dfrac{\ln^{2018}(1+x)}{x^{2019}} = \\ \\ =\lim\limits_{x\to 0}\dfrac{\ln(1+x^{2018})}{x^{2018}}\cdot \dfrac{1}{x}- \lim\limits_{x\to 0}\Big[\dfrac{\ln(1+x)}{x}}\Big]^{2018}\cdot \dfrac{1}{x} = \\ \\ = \lim\limits_{x\to 0}\Big(1\cdot \dfrac{1}{x}-1^{2018}\cdot \dfrac{1}{x}\Big) =\lim\limits_{x\to 0}\Big(\dfrac{1}{x}-\dfrac{1}{x}\Big) = 0[/tex]
[tex]\boxed{\lim\limits_{u\to 0}\dfrac{\ln(1+u)}{u} = 1}[/tex]
