Supremum and infimum set with two variables

Question

I am trying to conclude about the supremum and infimum of $A=\left\{{2013n + k \over n + 2013k}: n, k \in \mathbb{Z}, n, k \ge 10,000\right\}$. I transformed the equation:

$$\begin{align} &{n+2013k+2012n-2012k \over n+2013k}=1+2012{n-k \over n+2013k}=1+2012{n+2013k-2014k \over n+2013k}= \\ & = 1+2012\left(1-{2014k \over n+2013k}\right) \end{align}$$ Intuitively, if I choose a relatively small $k$ and a sufficiently large $n$, the last term should come close to $0$ and the whole expression close to $2013$. Alternatively, if I choose a relatively small $n$ and large $k$, the last term seems to be going to $\infty$ and the whole expression to $-\infty$. So, my guesswork is that we might have a supremum but possibly no infimum. Is this in the right ballpark? Mind that I don't want to use limits in any formal way here as they haven't been properly defined in the material I'm going over.

Answer 1

" if I choose a relatively small n and large k, the last term seems to be going to ∞" - this is wrong, $\frac{2014k}{n+2013k}\in(0,\frac {2014} {2013})$

So infimum should be $\frac{1}{2013}$

Answer 2

The last term doesn't go to $ \infty $, since $ \frac{2014k}{n + 2013k} < \frac{2014k}{2013k} = \frac{2014}{2013} $, so for such choice of $ k,n $ you get that your infimum might be $ 1 + 2012\cdot \frac{-1}{2013} = \frac{1}{2013}$, but you need to finish the proof