help with solving a limit with logarithms

Question

I am preparing for an exam and I am struggling with the following limit: $$ \lim_{x\to 0}\frac{\ln(1+x^{2018})-\ln^{2018}(1+x)}{x^{2019}}. $$

I tried the L'Hospital rule and i tried to form remarcable limits.

Answer 1

Replace $2018$ by $n$ and then note that the expression under limit can be written as $$\frac{\log (1+x^n)-\log^n(1+x)}{x^{n+1}}$$ which can be further expressed as $$\frac{\log(1+x^n)-x^n}{x^{2n}}\cdot x^{n-1}-\frac{\log^n(1+x)-x^n}{x^{n+1}}$$ The first term clearly tends to $0$ and the second term can be written as $$-\dfrac{\left(\dfrac{\log(1+x)}{x}\right)^n-1}{\dfrac{\log(1+x)}{x}-1}\cdot\frac{\log(1+x)-x}{x^2}$$ which tends to $-n(-1/2)=n/2=1009$ and that is the desired limit.

Answer 2

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