# 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?

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