Consider the function $f:X\to Y$ given by $$f(a,b)=\frac{(\lfloor\frac{ab}{a+1}\rfloor+1)(a+1)-ab}{a(b-1)-1},$$ where $a\in\mathbb{R}_{>0}$ and $b\in\mathbb{N}$ and we impose the conditions $a>\frac{2}{b-2}$ and $b>2$. Note that the notation $\lfloor x\rfloor$ gives the greatest integer that is less than $x$ (i.e., the floor function).
We interested to study the limit behavior of function $f$ as $b$ goes to positive infinity. We have $$\lim_{b\to\infty}f(a,b)=\lim_{b\to\infty}\frac{(\lfloor\frac{ab}{a+1}\rfloor+1)(a+1)-ab}{a(b-1)-1}.$$
If the fraction $\frac{ab}{a+1}$ is not an integer, then can we assume that $\lfloor\frac{ab}{a+1}\rfloor=\frac{ab}{a+1}-\varepsilon$, where $\varepsilon>0$?
If yes, then we can get $$\lim_{b\to\infty}\frac{(\frac{ab}{a+1}-\varepsilon+1)(a+1)-ab}{a(b-1)-1}=\lim_{b\to\infty}\frac{ab+(1-\varepsilon)(a+1)-ab}{a(b-1)-1}=\lim_{b\to\infty}\frac{(1-\varepsilon)(a+1)}{ab-a-1},$$ which is "equivalent" to the standard limit $\lim_{x\to\infty}\frac{1}{x}=0$. Thus we can conclude that $$\lim_{b\to\infty}f(a,b)=0$$
i) Is my method a valid one to take the limit of the floor function defined above?
ii) In order to be rigorous, do I have to show that the limit of $f$ exist? Is the $\epsilon-\delta$ technique the only way to show existence of a limit?
I am not sure how to show existence with the floor function. Any hints would be much appreciated!
For $(i)$ your method is valid, but to be more rigorous $(ii)$, you can use squeeze theorem and the fact that $x-1 \leq \lfloor x \rfloor \leq x$ so your limits is bounded between $\lim \limits_{b \to \infty} \frac{(\frac{a b}{a+1})(a+1)-a b}{a(b-1)-1}$ and $\lim \limits_{b \to \infty} \frac{(\frac{a b}{a+1}+1)(a+1)-a b}{a(b-1)-1}$ which both approach $0$ as $b \to \infty$ using your method, so the limit is zero.
For $(ii)$ i think you can use squeeze theorem and some limits arithmetic and to know some limits like $\lim \limits_{x \to \infty} \frac{1}{x}= 0$ would be enough