Question: $$\begin{cases}h(x)=1&x\in\mathbb{Q}\\h(x)=-1&x\text{ irrational}
\end{cases}$$

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> Find $\lim\_{x\to \sqrt{2}} h(x)$
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It looks very simple because $\sqrt{2}$ is irrational , so $\lim\_{x\to \sqrt{2}} h(x)= -1$.

Is it true ?
Thanks.

Answer: 
The limit does not exist.

To see this, note that given any $\delta > 0$ there exists a rational number $x \in (\sqrt{2}, \sqrt{2}+\delta)$ and an irrational number $y \in (\sqrt{2}, \sqrt{2}+\delta)$. For such values of $x$ and $y$, we have $h(x)-h(y) = 2$, and so taking $\varepsilon = 1$ in the (negation of the) definition of a limit shows that no limit exists.

Intuitively, the function takes both values $1$ and $-1$ arbitrarily close to $\sqrt{2}$, so it can't have a limit at $\sqrt{2}$.
It's actually not defined because the rationals and irrationals are dense in the real numbers.

thinking sequentially: you can approach $\sqrt{2}$ by either rationals or irrationals so the limit must be both -1 and 1, which is impossible.