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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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    This isn't really about "limit of functions," which is a separate topic. This is really about a standard "limit."2012-11-16

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$\lim\limits_{t\to {x_{0}}}{f(t)}=A$ 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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    Thx, that's very clear2012-11-17
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The definition is $f(t) \xrightarrow[t \to x^+]{}a$ if $\forall \epsilon >0 \ \exists \ \delta>0 \ \text{s.t. if} \ x

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. 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 Therefore $\not \exists \ a \in \mathbb{R} \ $ s.t. $|f(t)-a|<\frac{1}{2}$ whenever $x.

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    Yours are very clear that using the definition to explain.2012-11-17
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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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    This is a good place where it makes sense to say "$\mathbb{Q}$ is dense in $\mathbb{R}$."2012-12-15