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Let $f(x)=0$ for $x\in\mathbb{Z}$ and $1$ otherwise. Why $\forall x\in\mathbb{Z},\lim_{t\to x^{+}}f(t)=\lim_{t\to x^{-}}f(t)=1 $. Can it be explained by using the definition of limit? Would the result the same if $\mathbb{Z}$ is changed to $\mathbb{Q}$ ???

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You repeated $\lim_{t\to x^+}f(t)$ in there... –  Arkamis Nov 16 '12 at 17:53
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Wikipedia: One-sided limit. (Since you asked in comments to answers what $t\to x^+$ means. –  Martin Sleziak Nov 16 '12 at 18:10
    
I think the rigorous definition in that website explains it –  Mathematics Nov 16 '12 at 18:14
    
This isn't really about "limit of functions," which is a separate topic. This is really about a standard "limit." –  Thomas Andrews Nov 16 '12 at 18:15

3 Answers 3

up vote 2 down vote accepted

$\lim\limits_{t\to {x_{0}}}{f(t)}=A$ means that $$(\forall \varepsilon>0 )\;\;(\exists \delta>0): \;\; \forall t\in B'_\delta(x_{0})=(x_0-\delta;\,x_0+\delta)\setminus\{x_0\} \\ |f(t)-A|<\varepsilon.$$ If $\mathbb{Z}$ is changed to $\mathbb{Q},$ i.e. $x \in \mathbb{Q}$ limit does not exist. To prove this, consider two cases:

  1. Let $\{t_n\} \subset \mathbb{Q}$ be an arbitrary sequence of rationals converging to $x:\;\; t_n \underset{n \to{\infty}}{\to} x.$ Then $\lim\limits_{t=t_n\to {x_{0}}}{f(t)}=\lim\limits_{n\to {\infty}}{f(t_n)}=1.$
  2. For arbitrary sequence of irrational numbers $\{t_{n}^{*}\}\subset \mathbb{R}\setminus \mathbb{Q},\;\; t_{n}^{*} \underset{n \to{\infty}}{\to} x$ we have $\lim\limits_{t=t_{n}^{*}\to {x_{0}}}{f(t)}=\lim\limits_{n\to {\infty}}{f(t_{n}^{*})}=0.$
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i know what that mean, but in this case, it is $x^{+}$. What if $\mathbb{Z}$ is changed to $\mathbb{Q}$ –  Mathematics Nov 16 '12 at 18:06
    
Thx, that's very clear –  Mathematics Nov 17 '12 at 3:20

The definition is $f(t) \xrightarrow[t \to x^+]{}a$ if $\forall \epsilon >0 \ \exists \ \delta>0 \ \text{s.t. if} \ x<t<x+\delta \Rightarrow |f(t)-a|<\epsilon.$

For $\displaystyle{\lim_{t\to x^{+}}f(t)}=1, \ \text{where} \ x \in \mathbb{Z}$:
let $\epsilon>0$. Then for $\delta=\frac{1}{2}$ if $x<t<x+\delta=x+\frac{1}{2} \Rightarrow t \not \in \mathbb{Z} \Rightarrow f(t)-1=0$. Therefore $f(t) \xrightarrow[t \to x^+]{}1$.
Similar for $\displaystyle{\lim_{t\to x^{-}}f(t)}=1$.

If we replace $\mathbb{Z}$ with $\mathbb{Q}$ the $\displaystyle{\lim_{t\to x^+}f(t)}$ (or $\displaystyle{\lim_{t\to x^{-}}f(t)}$ or $\displaystyle{\lim_{t\to x}f(t)}$) does not exists. For, if $\epsilon = \frac{1}{2}$ then for every $\delta>0$ there are $t_1\in \mathbb{Q}, \ t_2\not \in \mathbb{Q}$ s.t. $x<t_1,t_2<x+\delta \Rightarrow f(t_1)=0 , \ f(t_2)=1.$ Therefore $\not \exists \ a \in \mathbb{R} \ $ s.t. $|f(t)-a|<\frac{1}{2}$ whenever $x<t<x+\delta \ \ \ \square$.

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Yours are very clear that using the definition to explain. –  Mathematics Nov 17 '12 at 3:20

Yes. Take any ball around $1$. Then whatever integer you choose, you can take a small enough ball around it so that anything but that integer maps to $1$ under $f$.

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Can you explain what exactly does $t\to x^{+} or t\to x^{-}$.Also what if $\mathbb{Z}$ is changed to $\mathbb{Q}$, will the result be the same ??? –  Mathematics Nov 16 '12 at 18:05
    
I assume you are using $\epsilon, \delta$ definition of limit. Answer to your first question: It's like splitting the $\delta$-ball into half and consider each piece. Answer to your first question: The result must be different if you use $\mathbb{Q}$ because you cannot take (half) $\delta$-ball so that everything in that set maps to only one of $0$ xor $1$. –  GYC Dec 15 '12 at 15:10
    
This is a good place where it makes sense to say "$\mathbb{Q}$ is dense in $\mathbb{R}$." –  GYC Dec 15 '12 at 15:17

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