Proving discontinuity and continuity of floor function

Question

I am trying to prove $\lim\limits_{x\to{a}} \lfloor{x}\rfloor=\lfloor{a}\rfloor$ if $x \not\in \mathbb{Z}$ and that the limit does not exist if it is in $\mathbb{Z}$. I believe there is some error in my $δ-ε$ argument.

Proof:
If $0<\bigl|{\lfloor{x}\rfloor-\lfloor{a}\rfloor}\bigr| < ε$, then $\bigl|\lfloor{x}\rfloor\bigr|-\bigl|\lfloor{a}\rfloor\bigr|\le \bigl|{\lfloor{x}\rfloor-\lfloor{a}\rfloor}\bigr| <ε$, so $$\bigl|\lfloor{x}\rfloor\bigr| < ε+\bigl|\lfloor{a}\rfloor\bigr|.$$ Now by definition, $\lfloor{x}\rfloor=x-\{x\}$, so $\bigl|x-\{x\}\bigr| < ε+\bigl|\lfloor{a}\rfloor\bigr|$. Thus, $$|x|-1<|x|-|{x}|\le\bigl|x-\{x\}\bigr| < ε+\bigl|\lfloor{a}\rfloor\bigr| \implies |x| < ε+\bigl|\lfloor{a}\rfloor\bigr|+1.$$ This in turn implies that if $0<|x-a|<ε+\lfloor{a}\rfloor-a+1=δ \implies 0<\bigl|{\lfloor{x}\rfloor-\lfloor{a}\rfloor}\bigr| < ε $ as desired. Our choice of $δ>0$, as $\lfloor{a}\rfloor-a+1>0$ as this implies $1>{a}$ which is trivially true.

However, I do not see where my argument fails if $x \in \mathbb{Z}$, and am thus confused. I know how to prove it is discontinuous using the $δ-ε$ definition, but I do not see how my proof "fails" in the case that $x$ is an integer.

Answer 1

As noted in the comments, the cases need to be based, not on whether or not $x\in\mathbb{Z}$, but rather, whether or not $a\in\mathbb{Z}$.

With that correction, the claims to be proved are:

• If $a\notin\mathbb{Z}$, then ${\displaystyle{\lim_{x\to a}\,\lfloor{x}\rfloor=\lfloor{a}\rfloor}}$.$\\[10pt]$
• If $a\in\mathbb{Z}$, then ${\displaystyle{\lim_{x\to a}\,\lfloor{x}\rfloor}}$ does not exist.

First suppose $a\notin\mathbb{Z}$.

Even ignoring the mixup of $x\notin\mathbb{Z}$ versus $a\notin\mathbb{Z}$, your proof attempt goes immediately off course . . .

To prove ${\displaystyle{\lim_{x\to a}\,\lfloor{x}\rfloor=\lfloor{a}\rfloor}}$ with an $\epsilon$-$\delta$ type proof, it's not correct to start with a sentence of the form

If $0 < \bigl|\lfloor{x}\rfloor - \lfloor{a}\rfloor\bigr| < \epsilon\,$, . . .

Rather, you should start with . . .

1. Fix $\epsilon > 0$.$\\[8pt]$
2. Let $\delta =\;$[some expression, possibly depending on $\epsilon]$.

and then try to show

If$\;0 < |x-a| < \delta$, then $\bigl|\lfloor{x}\rfloor - \lfloor{a}\rfloor\bigr| < \epsilon$.

Here's a proof along those lines (for the case $a\notin\mathbb{Z}$) . . .

Fix $\epsilon > 0$.

Let $\delta=\min(a-\lfloor{a}\rfloor,\lceil{a}\rceil-a)$.

Since $a$ is not an integer, it follows that $\delta > 0$.

Note that in this case, the expression chosen for $\delta$ is independent of the choice of $\epsilon$.

Our goal is to show that if $\;0 < |x-a| < \delta$, then $\bigl|\lfloor{x}\rfloor - \lfloor{a}\rfloor\bigr| < \epsilon$.

Thus, assume $0 < |x-a| < \delta$. \begin{align*} \text{Then:}\;\; x-\lfloor{a}\rfloor &= (x-a)+(a-\lfloor{a}\rfloor)\\[4pt] &> -|x-a| + \delta\\[4pt] &> -\delta + \delta\\[4pt] &= 0\\[16pt] \text{Also:}\;\; \lceil{a}\rceil-x &=(\lceil{a}\rceil-a)+(a-x)\\[4pt] &> \delta - |x-a|\\[4pt] &> \delta - \delta\\[4pt] &= 0\\[4pt] \end{align*} hence $\lfloor{a}\rfloor < x < \lceil{a}\rceil$.

Since $a\notin\mathbb{Z}$, it follows that $\lfloor{a}\rceil$ and $\lceil{a}\rceil$ are consecutive integers, hence \begin{align*} &\lfloor{a}\rfloor < x < \lceil{a}\rceil\\[4pt] \implies\;&\lfloor{x}\rfloor=\lfloor{a}\rfloor\\[4pt] \implies\;&\bigl|\lfloor{x}\rfloor-\lfloor{a}\rfloor\bigr|=0\\[4pt] \implies\;&\bigl|\lfloor{x}\rfloor-\lfloor{a}\rfloor\bigr|< \epsilon\\[4pt] \end{align*} as was to be shown.

Next, suppose $a\in\mathbb{Z}$.

Using arguments similar to the one given above, I'll leave it to you to show that

• ${\displaystyle{\lim_{x\to a^{-}}\,\lfloor{x}\rfloor=a-1}}$
  $\qquad$[Hint: Let $\delta=1$]$\\[10pt]$
• ${\displaystyle{\lim_{x\to a^{+}}\,\lfloor{x}\rfloor=a}}$
  $\qquad$[Hint: Let $\delta=1$]

hence conclude that ${\displaystyle{\lim_{x\to a}\,\,\lfloor{x}\rfloor}}$ does not exist.