Proof that Thomae's function doesn't have a limit as $\lim_{x\rightarrow x_0}$, $x_0\in \mathbb{Q}$ I'm trying to proof that the Thomae's function doesn't have a limit as  $\lim_{x\rightarrow x_0}$, $x_0\in \mathbb{Q}$.
I'm pretty sure, that I can solve this by using the $\epsilon-\delta$-critereon, but I cant figure out how. If a limit would exist, that would mean:
$\forall\epsilon>0 \exists\delta : |f(x)-L|<\epsilon, \forall x\in D, 0<|x_i -x|<\delta$
I'm pretty sure that you simply can't find a matching $\delta$ as in the environment of each rational number there are some irrational numbers where the Thomae's function is equal to $1/q$ and not $0$. 
Could you help me to complete this proof by helping me arguing&writing this mathematicly correct?
Edit:
This isn't a duplicate because I'm looking for the limit and not discontinuity. I have to proof this with the $\epsilon-\delta$-critereon for the limit and not for the discontinuity (even though they are kind of similar, they're different!)
 A: The comments by user Wojowu are correct and your exercise has a mistake which is not just a typo. When reading typical calculus books always exercise caution. Most of them have serious errors in their exercises as well as proofs.
For clarity I define the Thomae function as $f:[0, 1]\to\mathbb {R} $ by $f(x) =0$ if $x$ is irrational and $f(x) =1/q$ if $x=p/q$ is a rational number such that $p, q$ have no common factor.
Then for all $a\in[0,1]$ the limit $L_a=\lim_{x\to a} f(x)$ exists and is equal to $0$. The idea behind the proof is not difficult. Try to convince yourself of the following simple result.

Simple Theorem: Let $a, b\in\mathbb{R}, a<b$ and $N$ be a positive integer. Then there are only a finite number of rationals in interval $[a, b]$ whose denominators do not exceed $N$.

Now as $x\to a$ through irrational values we have $f(x) =0$. And if $x\to a$ through rational values the denominators $q$ of $x$ will increase without bound and hence $f(x) =1/q\to 0$. Thus $\lim_{x\to a} f(x) =0$.
A formal $\epsilon, \delta$ proof makes use of the simple theorem mentioned above.
Let's start with $\epsilon >0$ and choose a positive integer $N$ such that $1/N<\epsilon$. Now there are only a finite number of rationals in $[0,1]$ with denominators less than or equal to $N$. Out of these rationals choose the one which is nearest but not equal to $a$ and call it $b$. Let $\delta=|a-b|$ and consider all $x$ with $0<|x-a|<\delta$. If $x$ is irrational then $f(x) =0$ and if $x$ is rational then it has denominator $M$ greater than $N$ so that $f(x) =1/M<1/N<\epsilon $. Thus we have $|f(x) |<\epsilon $ whenever $0<|x-a|<\delta$. And hence $\lim_{x\to a} f(x) =0$. This also proves that the function $f$ is continuous at irrational points and discontinuous at rational points in $[0,1]$.
