$g(x) = \begin{cases} \frac{1}{b} \text{ if } x^2 = \frac{a}{b} \in \mathbb{Q} \text{ in lowest terms}\\ 0 \text{ if } x^2 \notin \mathbb{Q} \end{cases}$
Evaluate $\displaystyle{\lim_{x \to 0} g(x)}$
Here's what I've done to evaluate the limit. Is this correct/sufficient to show that the limit as $x$ approaches $0$ is $0$?
$\displaystyle{\lim_{x \to 0} 0 = 0}$
$\displaystyle{\lim_{x \to 0} x^2 = 0}$
$\forall x \in \mathbb{Q} \text{, } 0<x<1 \text{, } 0 < \frac{1}{b} \leq \frac{a}{b} = x^2$
$\Rightarrow$ by the Squeeze Theorem, the limit of $g(x)$ as $x \rightarrow 0^+$ is $0$
Edit: Fixed formatting
It's clear that you have the right idea, but you might want to be a bit more precise with your proof. A piece of advice I would offer is to use more English words. Mathematics isn't a competition to use as much notation as possible!
Here is how you might want to more clearly write the proof.
Without loss of generality, we may assume that $x \in (-1,1)$ as we are only considering the limit as $x \to 0$. Let $f(x) = x$. We see that if $x^2 = \frac{a}{b} \in \mathbb{Q}$, then $|g(x)| = |\frac{1}{b}| \leq |\frac{a}{b}| = |x^2| <|x|$ as $|x|<1$. If $x^2 \notin \mathbb{Q}$, then $|g(x)| = 0 <|x|$ and so $|g(x)| \leq |f(x)|$ for all $x \in (-1,1)$. It follows from the squeeze theorem that $\displaystyle{0 \leq \lim_{x \to 0} |g(x)| \leq \lim_{x \to 0} |f(x)| = 0}$ and so $$\lim_{x \to 0} |g(x)| = 0,$$ hence $\displaystyle{\lim_{x \to 0} g(x) = 0}$.