Is this graph of $f(x)=\frac{1}{\left [ \frac{1}{x} \right ]}$ present in Spivak's Calculus correct?

Question

The following problem is from Ch. 4 "Graphs" from Spivak's Calculus

17. The symbol $[x]$ denotes the largest integer which is $\leq x$. Thus, $[2.1]=[2]=2$ and $[-0.9]=[-1]=-1$. Draw the graph of the following functions (they are all quite interesting, and several will reappear frequently in other problems).

(vi) $f(x)=\frac{1}{\left [ \frac{1}{x} \right ]}$

The solution manual solution for this particular item is

Is this actually correct for values of $x\leq 0$?

For example consider the interval $[-1/2,-1/3]$. The graph above seems to imply that for any $x \in (-1/2,-1/3)$, we have $f(x)=f(-1/2)$.

But consider one such value, for example $-0.4=-\frac{2}{5}$. Then

$$\left [ \frac{1}{-2/5} \right ]= \left [ -\frac{5}{2} \right ]= [-2.5]=-3$$

Hence

$$f(-2/5)= -\frac{1}{3}=f(-1/3)$$

Second of all, why is the function not defined for $x\leq -1$?

For example,

$$f(-2)=\frac{1}{\left [ \frac{1}{-2} \right ]}=\frac{1}{-1}=-1$$

$$f(-1000)=\frac{1}{\left [ \frac{1}{-1000} \right ]}=\frac{1}{-1}=-1$$

Shouldn't the graph look like

EDIT:

Note that the solution manual also has the following graph

for the function

$$f(x)=\left [ \frac{1}{x} \right ]$$

which also seems to be incorrect.

Answer 1

Good catch!

It seems that the author is using the integer part function $\operatorname{int}(x)$ which always rounds towards zero rather than the floor function $\lfloor x \rfloor$ which always rounds down. They agree for all $x \geq 0$, but for $x < 0$,
$$ \operatorname{int}(x) = \lceil x \rceil = - \lfloor -x \rfloor $$ If $x \notin \mathbb{Z}$ and $x<0$, the difference is precisely one: $$ \operatorname{int}(x) = 1 + \lfloor x \rfloor, $$ but the two functions do agree at all integer values.

Another distinction, which is visible in the graphs: the floor function, is always continuous from the right $$ \lim_{x \to a^+} \lfloor x \rfloor = \lfloor a \rfloor, $$ whereas the integer part function is always continuous from the direction away from zero. Hence, for $a<0$, $$ \lim_{x \to a^-} \operatorname{int}(x) = \operatorname{int}(a). $$

You can compare them here by clicking on one or the other folder to toggle the graphs on/off.