Limit infinity involving floor function

Question

I'm studying calculus right now but I'm stuck at solving a limit involving floor function.

The problem is to find

$\lim_{x \to 1^{-}}(\dfrac{\lfloor x^2 \rfloor - (\lfloor x \rfloor)^2}{x^2-1})=-\infty$

My first thought was to let $x<1 \Rightarrow \lfloor x^2 \rfloor = 0$, $x<1 \Rightarrow \lfloor x \rfloor = 0$ so when $x \to 1^{-} $ then $lim_{x \to 1^{-}}=\dfrac{0}{0}$. But I can't go any further and don't know whether my thought is correct.

Answer 1

When $0\leq x <1$ we have that $\lfloor x \rfloor = \lfloor x^2\rfloor = 0$, so $$ \lim_{x\to 1^-}\dfrac{\lfloor x^2 \rfloor - \lfloor x \rfloor^2}{x^2-1} = \lim_{x\to 1^-}\dfrac{0}{x^2-1} = 0 $$

Answer 2

You are almost there. Note that for all $x\in(1/2,1)$, as you have noted, both $[x^{2}]=0$ and $[x]=0$, and so for these $x$, $1/(x^{2}-1)$ is defined and we have $([x^{2}]-[x]^{2})/(x^{2}-1)=0$. We conclude that the limit is zero.