Left and Right limits of $\lfloor\lfloor x\rfloor\rfloor$ at $x = 0$

Question

I'm self-studying calculus from Larson's Calculus 8E and on page 102 and I don't understand why

$$\lim_{x\to 0^-} \frac{f(x) - f(0)}{x-0} = \frac{\lfloor\lfloor x\rfloor\rfloor - 0}{x} = \infty $$

and

$$\lim_{x\to 0^+} \frac{f(x) - f(0)}{x-0} = \frac{\lfloor\lfloor x\rfloor\rfloor- 0}{x} = 0 $$

Why isn't the first limit 0? Why is the second limit not $\infty$?

Answer 1

Well, $\frac{f(x)-f(0)}{x-0}=\frac{-1}{x}$ for $x<0$ and $\frac{f(x)-f(0)}{x-0}=0$ for $x>0$.

Can you take it from here?

Answer 2

For the first, note that for $-1\lt x\lt 0$, $\lfloor x\rfloor=-1$, so $\lfloor\lfloor x\rfloor\rfloor=-1$.

For the second, note that for $0\lt x\lt 1$, $\lfloor x\rfloor=0$, so $\lfloor\lfloor x\rfloor\rfloor=0$.