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In my Calculus home work assignment I get the following tricky question about the limit of following function:

Define a function $f : mathbb{R} rightarrow mathbb{R}$ as follows:

• if $x in mathbb{R}setminusmathbb{Q}$, then $f(x) = 0$.


• if $x in mathbb{Q}$, let $p in mathbb{Z}$, $q in mathbb{N}$ such that $x = frac{p}{q}$ is the reduced form of $x$. Then $f(x) = frac{1}{q}$.

Prove that $lim_{x rightarrow x_0} f(x) = 0$ for every $x_0 in mathbb{R}$.

Hint: You may rely on the following claim:

Let $k,d > 0$ be positive real numbers, and let $x_0 in mathbb{R}$. There is finite number of integers $p in mathbb{Z}$, $q in mathbb{N}$, such that $q leq k$ and $ left| x_0 - frac{p}{q} right| < d$.

I'm struggling with this question 3 days already. Intuitively I understand that for every $epsilon > 0$ I can choose $delta > 0$ small enough, such that $frac{1}{q} < epsilon$, but I'm really struggling to put condition on $delta$.

Consider the number $n!$ (for some given $n in Bbb N$).

For each number $x_0 in Bbb R$, there must exist $a in Bbb Z$ such that

$$frac{a}{n!}<x_0 leq frac{a+1}{n!}$$

Note that $a$ depends on both $x_0$ and $n$.

Also, observe that $f(x)<frac 1n$ in each such interval $big(frac{a}{n!} , frac{a+1}{n!} big)$. Thus,

begin{align}
& forall varepsilon >0 ; ; exists ; delta = min bigg { bigg| x_0 - frac{a}{big(lceilfrac 1varepsilon rceil big)!} bigg| ;, ; bigg| x_0 - frac{a+1}{big(lceilfrac 1varepsilon rceil big)!} bigg| bigg } >0 ; text{ such that } \
; |x-x_0|<delta & implies frac{a}{big(lceilfrac 1varepsilon rceil big)!} < x< frac{a+1}{big(lceilfrac 1varepsilon rceil big)!} \
& implies |f(x)-0| = f(x)<frac{1}{lceil frac 1varepsilon rceil} leq varepsilon
end{align}