Answer to Question #315582 in Calculus for Aune

Question

Let 𝑓(𝑥) = ⌊𝑥⌋ + ⌊−𝑥⌋, where ⌊𝑥⌋ is the greatest integer less than or equal to 𝑥.

(𝑎) For what values of 𝑎, does lim𝑥→𝑎

𝑓(𝑥)exist?

(𝑏) At what numbers is 𝑓 discontinuous?

Answer 1

(a) Let $a∈ℝ$ such that $a∉ℤ$ .

let $\lfloor{a}\rfloor=m$

$=> \lfloor{-a}\rfloor=-(m+1)$

There exist $ε>0$ such that $∀ r∈(a-ε,a+ε)$

$=> \lfloor{r}\rfloor=m$

and $\lfloor{-r}\rfloor=-(m+1)$

Thus,

$\lim\limits_{x\rarr a^{+}}f(x)=\lim\limits_{x\rarr a^{-}} f(x)=\lim\limits_{x\rarr a}f(x)$

$=\lfloor{r}\rfloor+\lfloor{-r}\rfloor=m-(m+1)=-1$

Suppose $a∈ℤ$ . There exist $ε>0$ such that $∀ k∈(a-ε, a)$

$=> \lfloor{k}\rfloor=a-1$

and $\lfloor{-k}\rfloor=-a$

$\lim\limits_{x\rarr a^{-}}f(x)=\lfloor{k}\rfloor+\lfloor{-k}\rfloor=(a-1)-a=-1$

Also, there exists $ε>0$ such that $∀ k∈(a, a+ε)$

$=> \lfloor{k}\rfloor=a$

and $\lfloor{-k}\rfloor=-(a+1)$

$\lim\limits_{x\rarr a^{+}}f(x)=\lfloor{k}\rfloor+\lfloor{-k}\rfloor=a-(a+1)=-1$

$\lim\limits_{x\rarr a^{+}}f(x)=\lim\limits_{x\rarr a^{-}} f(x)=\lim\limits_{x\rarr a}f(x)=-1$

Hence, $\lim\limits_{x\rarr a}f(x)$ exists for all $a∈ℝ$

(b) Let $x∈ℝ$ such that $x∉ℤ$

let $\lfloor{x}\rfloor=m$

$=>\lfloor{-x}\rfloor=-(m+1)$

$f(x)=\lfloor{x}\rfloor+\lfloor{-x}\rfloor=m-(m+1)=-1$

We established in (a) above that $\lim\limits_{x\rarr a}f(x)=-1$ , where a is any non-integer

Thus, $\lim\limits_{x\rarr a}f(x)=f(x)$ .

In addition, $f(x)$ is defined $∀ x∈ℝ$

Thus, f is continuous at any non-integer point

Suppose $x∈ℤ$ .

$f(x)=\lfloor{x}\rfloor+\lfloor{-x}\rfloor=x-x=0$

We established in (a) above that $\lim\limits_{x\rarr a}f(x)=-1$ , where a is any integer

Thus, $\lim\limits_{x\rarr a}f(x)≠f(x)$

Hence, f is discontinuous at any integer point.

f is discontinuous $∀x∈ℤ$