Assume we have a sequence of rational numbers $\left(\frac{p_n}{q_n}\right),$ where $\gcd(p_n,q_n)=1, \ \forall n \in \mathbb N$.
We know that $$\lim_{n\to\infty} \left(\frac{p_n}{q_n}\right)= x$$ and $$\lim_{n\to\infty} p_n = \lim_{n\to\infty} q_n = \infty.$$
Assume $x = \frac{M}{N},$ where $\gcd(M,N)=1$. Now, there exists such $k \in \mathbb N$ that $p_n>M$ and $q_n > N, \ \forall n\ge k$.
Does this prove that x is irrational?
On
Given $x = M/N$ as you say, good approximations parametrized by $n \in \mathbb N$ include $$ \frac{nM + 1}{nN + 1} $$
A (real) number is irrational if and only if the simple continued fraction for it is infinite. If the number is a "quadratic irrational" $(A + \sqrt B)/ C$ For integers $A,B,C,$ with $C \neq 0$ and $B$ positive but not a square, then the s.c.f. is eventually periodic and infinite.
If $x = M/N$ with positive integers $M,N,m,n$, either $|x - m/n| = 0$ or $|x - m/n| \ge \dfrac{1}{Nn}$. On the other hand, if $x$ is irrational there are infinitely many pairs $(m,n)$ with $0 < |x - m/n| < \dfrac{1}{n^2}$. In this sense, irrational numbers can be very well approximated by rationals, but rational numbers can't be very well approximated by other rationals.