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.
Compute $\lim \limits_{x\to 0}\frac{\ln(1+x^{2018} )-\ln^{2018} (1+x)}{x^{2019} }$
limits
Related Solutions
Here is another approach which is somewhat simpler than the one given in another answer here.
I establish that $$\int_{0}^{1}f(x)\{nx\}\,dx\to\frac{1}{2}\int_{0}^{1}f(x)\,dx$$ as $n\to\infty $. The integral on left of above equation can be split as sum of $n$ integrals $$\sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n}f(x)\{nx\}\,dx=\frac{1}{n}\sum_{k=0}^{n-1}\int_{k}^{k+1}f(t/n)\{t\}\,dt$$ Using mean value theorem for integrals the right hand side of the above equation can be written as $$\frac{1}{n}\sum_{k=0}^{n-1}f(t_k/n)\int_{k}^{k+1}\{t\}\,dt$$ where $t_k\in[k,k+1]$ and since $\{t\} $ is periodic with period $1$ the above reduces to $$\left(\int_{0}^{1}\{t\}\,dt\right)\cdot\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{t_k}{n}\right)$$ The integral above is $1/2$ as $\{t\} =t$ if $t\in[0,1)$ and the next factor is Riemann sum for $f$ or $[0,1]$. Thus the above tends to $$\frac{1}{2}\int_{0}^{1}f(x)\,dx$$ Above derivation assume that $f$ is continuous on $[0,1]$. Putting $f(x) =x^{2019}$ we get the desired limit as $1/4040$.
More generally we can use same method to prove that $$\lim_{n\to\infty} \int_{0}^{1}f(x)g(\{nx\})\,dx=\left(\int_{0}^{1}f(x)\,dx\right)\left(\int_{0}^{1}g(x)\,dx\right)$$ where $f$ is continuous on $[0,1]$ and $g$ is of constant sign and Riemann integrable on $[0,1]$.
Going further we can also note that if $g$ is periodic with period $T$ and of constant sign and Riemann integrable on $[0,T]$ and $f$ is continuous on $[0,T]$ then $$\lim_{n\to\infty} \int_{0}^{T}f(x)g(nx)\,dx=\frac{1}{T}\left(\int_{0}^{T}f(x)\,dx\right)\left(\int_{0}^{T}g(x)\,dx\right)$$
Based on suggestion in comments, one can prove that the above result holds for Riemann integrable $f, g$ and $g$ also being periodic with period $T$.
The idea is to express the integral on left as a sum $$\frac{1}{n}\sum_{k=0}^{n-1}\int_{kT}^{(k+1)T}f(x/n)g(x)\,dx$$ which can be further rewritten as $$\frac{1}{n}\sum_{k=0}^{n-1}\int_{0}^{T}f((x+kT)/n)g(x+kT)\,dx$$ And since $g$ is periodic it follows that the above can be written as $$\frac{1}{T}\int_{0}^{T}\left(\frac{T}{n}\sum_{k=0}^{n-1}f\left(\frac{x+kT}{n}\right)g(x)\right)\,dx\tag{1}$$ Since $f$ is Riemann integrable on $[0,T]$ with integral $I=\int_{0}^{T}f(x)\,dx$ we can see that if $$P_n=\{0,T/n,2T/n,\dots,(n-1)T/n,T\} $$ is a partition of $[0,T]$ and $U(f, P_n), L(f, P_n) $ be corresponding upper and lower Darboux sums then we have $$L(f, P_n) \leq S(f, P_n) \leq U(f, P_n)$$ where $S(f, P_n) $ is any Riemann sum for $f$ over $P_n$. Since the integral $I$ is also sandwiched between both upper and lower sums we have $$|S(f, P_n) - I|\leq U(f, P_n) - L(f, P_n) $$ We can now observe that the integrand in equation $(1)$ is of the form $S(f, P_n) g(x) $ and hence $$\left|\int_{0}^{T}S(f,P_n)g(x)\,dx-I\int_{0}^{T}g(x)\,dx\right|\leq (U(f, P_n) - L(f, P_n)) \int_{0}^{T}|g(x)|\,dx$$ and clearly the right hand side above tends to $0$ so that the left hand side also does the same. It follows that the desired limit is $$\frac{1}{T}\int_{0}^{T}f(x)\,dx\int_{0}^{T}g(x)\,dx$$ Credit for the idea of above proof must go to the user WE Tutorial School.
If the integral $\int_{0}^{T}g(x)\,dx=0$ then the above can be used as a proof of Riemann-Lebesgue Lemma for Riemann integrable functions and therefore the above is a generalization of it.
Best Answer
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 $n\gt 1$.