I do not believe a reliable reference can be found since the method you describe is inherently flawed. The value $n$ is *never* "equal to" $\infty$. Likewise, there is no such number as "$(x/\infty)$," and it is most certainly not equal to 0.

Rather, it would appear that you are attempting to evaluate the limit of the series
$$
\sum_{i=1}^{n}\frac{1}{n}
$$
as $n$ approaches $\infty$ by treating the series as

$$
\lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{n} = \sum_{i=1}^{n}\lim_{n\to\infty}\frac{1}{n} = \underbrace{\lim_{n\to\infty}\frac{1}{n} +\lim_{n\to\infty}\frac{1}{n} + \cdots + \lim_{n\to\infty}\frac{1}{n}}_{n\ \text{times}}
$$
which is nonsensical, since the number of elements in the series depend the value of $n$, which no longer "exists" outside of the sum.

You may find the answers to this semi-related question helpful, which concludes that $1=2$ by computing $$\frac{\mathrm d}{\mathrm d x}x^2 = \frac{\mathrm d}{\mathrm d x} \underbrace{x+x+\cdots+x}_{x \ \text{times}}$$

As noted in the comments, this is mathematics, not physics, and so we do not have to worry about physical limitations such as "going below the plank scale." Real numbers are not quantized, and we are free to talk about finite values that are as small as we like.