Limit Question on $\mathbb{Q} \cap (0,1)$ Suppose that for $m,n$ that are coprime $f(m/n) = 1/(m+n)$. I want to show that the limit of $f(x) = 0$ as $x\to 0$ where the set of $x = \mathbb{Q} \cap  (0,1)$.
Clearly this is intuitively obvious. Consider the sequence $(x) = (1/2, 1/3, 1/4, \ldots)$ then you get $(1/3, 1/4, 1/5, \ldots)$ etc.
I'm trying to generalize this but am trying to see how the coprime plays into it? 
 A: Here's a hint: in a given $\varepsilon$ neighborhood of $0$, $(0, \varepsilon)$, how many fractions of the form $\frac{m}{2}$ are there? How many fractions of the form $\frac{m}{3}$, and so on? Based on the answer to that question, you can say something about the denominators of the elements of any sequence that converges to zero.
A: For $\epsilon >0$ take $A\in \Bbb N$ such that $\epsilon > \frac {1}{A+1}.$  Let $\delta=\frac {1}{A+1}.$ 
For all $x\in \Bbb Q\cap (0,\delta)$ we have $f(x)\in (0,\epsilon).$ 
Because if $x=\frac {m}{n}$ with co-prime $m,n\in \Bbb N$ then $$f(x)\geq \epsilon \implies m+n\leq \frac {1}{\epsilon}<A+1\implies n< A\implies x=\frac {m}{n}\geq \frac {1}{n}>\frac {1}{A}>\delta.$$
The idea is that if  $A\in \Bbb N$ then the smallest $x\in \Bbb Q^+$ such that $f(x)\geq \frac {1}{A+1}$ should be  $x=\frac {1}{A }$, so if $x\in \Bbb Q\cap (0,\frac {1}{A+1})$ then $f(x)<\frac {1}{A+1}.$ The key "move" is that $\frac {m}{n}\geq \frac {1}{n}.$
As noted in the comment from Bungo we require $m,n$ co-prime so that $f$ is well-defined. E.g. $\frac {1}{2}=\frac {2}{4}$  and it is illogical to have $1/(1+2)=f(1/2)=f(2/4)=1/(2+4).$
