irrational numbers - How can I prove that no limit exists for this function over this particular interval?

I was given the following function: $$f(x) = \begin{cases} \frac{1}{q} &\text{if }x = \frac{p}{q} \text{ is rational, reduced to lowest terms} \\ 0 &\text{if }x \text{ is irrational}\end{cases}$$
And was asked to attempt to prove or disprove whether or not, for some $j$ on the interval $(0,1)$, $\lim \limits_{x \to j}{}f(x)$ exists.
Intuitively, I know that I can find an irrational number between any two rationals which would obviously result in some discontinuity. I considered the idea that maybe it could be possible assuming that among the infinitely many rationals on the interval which also share a common denominator and could be used to find the limit, but that obviously would not work. However, I cannot formulate any proper line of reason to prove the limit does not exist.
How can I prove this, as formally as possible?

Answer

Recall the definition of $\lim_{x \to j}f(x)=a$:

$$\mbox{Given}\ \varepsilon > 0\ \mbox{there exists}\ \delta > 0\\ \mbox{such that}\ \left| f(x) - a\right|<\varepsilon\\ \mbox{whenever}\ 0<\left| x-j \right| <\delta$$

Now for any $j$, given $\varepsilon>0$, if we choose $\delta$ small enough we can ensure that the denominator $q$ of any fraction $\frac{p}{q}$ (in its lowest terms) in the interval $\left(j-\delta,j+\delta\right)$ satisfies $q>\frac{1}{\varepsilon}$, except possibly for $j$ itself, if it is rational. So that $0\le\left| f(x)\right|<\varepsilon$ whenever $0<\left|x-j\right|<\delta$.

Plugging this into the definition of a limit we see:

$$\lim_{x\to j}f(x) = 0\ \mbox{for all}\ j\in\left(0,1\right)$$

Now a function $g$ is continuous at $j$ iff $\lim_{x\to j}g(x)=g(j)$. It follows that $f$ is discontinuous at rational points and continuous at irrational points.