The limit equals $m/2$ when $m>1$ or $k\geq 1$, and it equals $m$ when $m=1$ and $k=0$. I provide the proof for $m>1$ below, the other cases being very similar.
Assume that $m>1$. The integral equals $m J_{m,k}$, where $$J_{m,k}(n):=\int_0^{1/m}(1-mu)^k\cot(\pi u)\sin(2\pi n u)\,du,$$ hence it suffices to show that $J_{m,k}(n)\to 1/2$ as $n\to\infty$. We observe that $$J_{m,0}(n)-J_{m,k}(n)=\int_0^{1/m}f_{m,k}(u)\sin(2\pi n u)\,du,$$ where $$f_{m,k}(u):=\left(1-(1-mu)^k\right)\cot(\pi u)$$ is continuous and bounded on $(0,1/m)$. Therefore, by basic Fourier analysis, the limit of $J_{m,0}(n)-J_{m,k}(n)$ is zero, hence it suffices to show that $J_{m,0}(n)\to 1/2$. Let us decompose $J_{m,0}(n)$ as $$J_{m,k}(n)=\int_0^{1/2}\cot(\pi u)\sin(2\pi n u)\,du-\int_{1/m}^{1/2}\cot(\pi u)\sin(2\pi n u)\,du.$$ As $m>1$, the function $\cot(\pi u)$ is continuous and bounded between $1/m$ and $1/2$, therefore the second integral tends to zero by basic Fourier analysis. Now we observe that the first integral is constant $1/2$, which finishes the proof. Indeed, \begin{align*} \int_0^{1/2}\cot(\pi u)\sin(2\pi n u)\,du&=\int_0^{1/2}\cos(\pi u)\sum_{k=1}^n 2\cos(2\pi(2k-1)u))\,du\\[8pt] &=2\sum_{k=1}^n\int_0^{1/2}\cos(\pi u)\cos(2\pi(2k-1)u))\,du. \end{align*} On the right hand side, the first term equals $1/4$, while all the other terms are zero, hence the left hand side equals $1/2$ as claimed.