Compute: $\lim\limits_{n\to+\infty}\int\limits_{0}^1 e^{\{nx\}}x^{100}dx$
We have that $e^{\{nx\}}$ is a $\frac{1}{n}$-periodic function and its mean value over a period is given by $(e-1)$.
By the Riemann-Lebesgue lemma (think to the Fourier series of $e^{\{x\}}$) it follows that:
$$ \lim_{n\to +\infty}\int_{0}^{1}e^{\{nx\}}x^{100}\,dx = (e-1)\int_{0}^{1}x^{100}\,dx = \color{red}{\frac{e-1}{101}}.$$
We do not really need it, but the explicit Fourier series of $e^{\{x\}}$ is given by:
$$ e^{\{x\}}=(e-1)+2(e-1)\sum_{m\geq 1}\frac{\cos(2\pi m x)-2m\pi \sin(2\pi m x)}{1+4m^2 \pi^2}$$
and the $m$-th term of the series is bounded by $\frac{1}{\sqrt{1+4m^2\pi^2}}$ due to the Cauchy-Schwarz inequality.
Additionally, for any $h\in\mathbb{N}^+$ we have
$$ \left|\int_{0}^{1}x^h \sin(2\pi n x)\,dx\right|\sim \left|\int_{0}^{1}x^h \cos(2\pi n x)\,dx\right|\sim\frac{1}{\pi n} $$
for large $n\in\mathbb{N}$.
An alternative approach. We have $$I=\int_{0}^{1}e^{\left\{ xn\right\} }x^{100}dx=\sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n}e^{\left\{ xn\right\} }x^{100}dx $$ $$\stackrel{y=nx}{=}\frac{1}{n}\sum_{k=0}^{n-1}\int_{k}^{(k+1)}e^{\left\{ y\right\} }\left(\frac{y}{n}\right){}^{100}dy $$ then by the mean value theorem exists some $z_{k}\in\left[k,k+1\right] $ such that $$I=\frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{z_{k}}{n}\right)^{100}\int_{k}^{(k+1)}e^{\left\{ y\right\} }dy $$ and since $e^{\left\{ x\right\} } $ has period $1$ $$ I=\frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{z_{k}}{n}\right)^{100}\int_{0}^{1}e^{y}dy=\left(e-1\right)\frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{z_{k}}{n}\right)^{100} $$ now if we take the limit we have a Riemann sum, so $$\begin{align} \lim_{n\rightarrow\infty}\int_{0}^{1}e^{\left\{ xn\right\} }x^{100}dx= & \left(e-1\right)\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{z_{k}}{n}\right)^{100} \\ = & \left(e-1\right)\int_{0}^{1}x^{100}dx=\color{red}{\frac{e-1}{101}}. \end{align}$$ Addendum: I realized that this approach works for any periodic function with period $1$. So it is easily generalizable.