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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.

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  • $\begingroup$ It should be that $a$ is not an integer, rather than $x$. $\endgroup$ – John Samples Jun 17 '18 at 2:35
  • $\begingroup$ You've probably misstated the claim. The correct cases are $a\notin\mathbb{Z}$, and $a\in\mathbb{Z}$. The variable $x$ has no choice of type (other than being real); it's forced to approach $a$ (which is constant). $\endgroup$ – quasi Jun 17 '18 at 2:36
  • $\begingroup$ The graph of the function $f(x) = \lfloor{x}\rfloor$ can guide you to the steps needed to prove the claims. If $a$ is a fixed integer, how does $f(x)$ behave as $x$ approaches $a$ from each side? Same question if $a$ is not an integer. $\endgroup$ – quasi Jun 17 '18 at 2:45
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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.

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  • $\begingroup$ Hello, I believe what I'm trying to say in my post is that if $\delta= \epsilon+\lfloor{a}\rfloor-a+1$ that it should imply that $0< |\lfoor{x}\rfloor-\lfoor{a}\rfloor}|< \epsilon$ and thus the result should follow. The part above that is not a "proof" but a derivation of a possible $\delta$. $\endgroup$ – Brian Jeesang Lee Jun 18 '18 at 22:40
  • $\begingroup$ @Brian Jeesang Lee: Ok, but then you should state that so the reader knows it's just side work, not part of the proof, and you should indicate where the proof starts. As it stands, it's not easy to see where your side work ends and where your proof begins. And as I indicated in my answer, a standard $\epsilon$-$\delta$ proof typically starts with $(1)\;$Let $\epsilon > 0$.$\;(2)\;$Let $\delta = $ . . . $\endgroup$ – quasi Jun 18 '18 at 22:52
  • $\begingroup$ @Brian Jeesang Lee: Also, for the case $a\notin\mathbb{Z}$, your proposed $\delta$, namely $\delta= \epsilon+\lfloor{a}\rfloor-a+1$, doesn't work. For example, let $\epsilon=\frac{1}{2}, a=\frac{3}{4}, x=\frac{5}{4}$. Then you get $$ \delta={\small{\frac{3}{4}}}\\ |x-a| = {\small{\frac{1}{2}}} < \delta\\ \text{but}\;\bigl|\lfloor{x}\rfloor-\lfloor{a}\rfloor\bigr| = 1 $$ which is not less than $\epsilon$. $\endgroup$ – quasi Jun 18 '18 at 23:16

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