As we have:
$$\sin^2(t)=\frac12(1-\cos(2t))$$
It is clear that the function being integrated has a periodicity of $\pi$. Hence every integral through a whole $\pi$ period will have the same value. So, we have that:
$$\int\limits_0^\pi\sin^2(t)dt=\int\limits_0^\pi\frac12(1-\cos(2t))dt=\frac\pi2$$
Then, if we break down the integration as the function is suggesting:
$$\int\limits_0^n\sin^2(t)dt=\int\limits_0^{\pi \lfloor\frac{n}{\pi}\rfloor}\sin^2(t)dt + \int\limits_{\pi \lfloor\frac{n}{\pi}\rfloor}^n\sin^2(t)dt=\frac\pi2 \lfloor\frac{n}\pi\rfloor + \int\limits_{\pi \lfloor\frac{n}{\pi}\rfloor}^n\sin^2(t)dt$$
Notice that the integrand is always positive, so we can easily find an lower and upper bound by excluding the left term or letting it complete another cycle, which means that:
$$\frac\pi2 \lfloor\frac{n}\pi\rfloor<\frac\pi2 \lfloor\frac{n}\pi\rfloor + \int\limits_{\pi \lfloor\frac{n}{\pi}\rfloor}^n\sin^2(t)dt<\frac\pi2 \lfloor\frac{n}\pi\rfloor+\frac\pi2$$
$$\frac1n \frac\pi2 \lfloor\frac{n}\pi\rfloor<U_n<\frac1n\left(\frac\pi2 \lfloor\frac{n}\pi\rfloor+\frac\pi2 \right)$$
So, if we let $n \to \infty$, as both sides of the inequality tend to $1/2$, we have that
$$\lim_{n\to\infty}U_n=\frac12$$
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 $n\gt 1$.
Best Answer
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.