Limit of argmin

Question

Fix $t \in [0,1]$ and $n \in \mathbb{N}$. Consider, for $r \in \mathbb{N}$, the ratio $\frac{i}{n+m}$ where $i \in I=\{0,1,...,n+r\}$. How can i prove that $$\lim_\limits{r \to \infty}\frac{i(r)}{n+r}=t$$ where $i(r)=\underset{i \in I}{\mathrm{argmin}}(\lvert t-\frac{i}{n+r}\rvert)$.

Answer 1

By construction $$ \Big| t - \frac{i(r)}{n+r} \Big| = \min_{i \in I} \Big| t - \frac{i}{n+r} \Big| \leq \Big| t - \frac{\lfloor t (n+r) \rfloor }{n+r} \Big| \leq \frac 1 {n+r} \underset{r \to \infty}{\longrightarrow} 0 $$