$$\lim_{x \to 0} \frac{\lfloor{x^2}\rfloor}{x^2}$$
$\lfloor x^2 \rfloor = 0 \space \forall x \in (-1,1)$ and $x^2 > 0 \space \forall \space x$ in the vicinity of $0$ but not at $0$. Hence this limit should be $\lim_{x \to 0} \frac{0}{x^2} = 0$.
The answer key in my textbook and GeoGebra, however, seem to disagree with this. Both claim that the limit is undefined. Why is this so?
This limit exists and is equal to zero. Consider any sequence $\{x_n\}_{n\in\Bbb N}$ converging to zero, none of whose terms is zero. Then, $$f(x_n)=\dfrac{\lfloor x_n^2\rfloor}{x_n^2}=0\,\forall n>N$$ where $N$ is a number such that $|x_n|<1\forall n>N$. Therefore, given limit is zero by sequential criteria.