Compute $\lim \limits_{x\to 0}\frac{\ln(1+x^{2018} )-\ln^{2018} (1+x)}{x^{2019} }$ I have to compute $\lim \limits_{x\to 0}\frac{\ln(1+x^{2018} )-\ln^{2018} (1+x)}{x^{2019}} $. I tried to use L'Hopital's Rule, but it didn't work. I also tried to divide both the numerator and denominator by $x^{2018} $, but I still have a $\frac{0}{0}$ indeterminate form. 
 A: By using the Maclaurin series of both terms the limit becomes
$$\lim_{x \to 0} \frac{x^{2018}+o(x^{4036}) - (x^{2018}-1009x^{2019}+o(x^{2020}))}{x^{2019}}$$
$$=\lim_{x \to 0} \frac{1009x^{2019}+o(x^{2020})}{x^{2019}}$$
$$=\lim_{x \to 0}( 1009+o(x))$$
$$=1009$$
In fact, one can generalise this limit giving the following solution
$$\lim_{x \to 0} \frac{\ln{(1+x^n)}-\ln^n{(1+x)}}{x^{n+1}}=\frac{n}{2}$$
for all $n\gt 1$.
A: The use of large exponents like $2018$ is more likely to remind the year in which the question was generated (or perhaps to intimidate the reader). It is best to replace it by a generic symbol $n$.
Thus we are interested in finding the value $f(2018) $ where $$f(n) =\lim_{x\to 0}\frac{\log(1+x^n)-(\log(1+x))^n}{x^{n+1}},n>1\tag{1}$$ Let's subtract and add $x^n$ in numerator and split the expression under limit as $$\frac{\log(1+x^n)-x^n}{x^{n+1}}+\frac{x^n-(\log(1+x))^n}{x^{n+1}}\tag{2}$$ The limit of both these terms can be evaluated on the basis of limit $$\lim_{t\to 0}\frac {\log(1+t)-t}{t^2}=-\frac{1}{2}\tag{3}$$ The above is not a standard / well known limit but can be easily evaluated using L'Hospital's Rule or Taylor series.
The first term in $(2)$ can be written as $$\frac{\log(1+x^n)-x^n}{x^{2n}}\cdot x^{n-1}$$ which tends to $(-1/2)\cdot 0=0$ via $(3)$. The second term in $(2)$ can be written as $$\dfrac{1-\left(\dfrac{\log(1+x)}{x}\right)^n}{1-\dfrac{\log(1+x)}{x}}\cdot\frac{x-\log(1+x)}{x^2}$$ which tends to $(n) (1/2)=n/2$ so that $f(n) =n/2$ and the desired limit $f(2018)$ is $1009$.
