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$f(x) = \begin{cases} x^2-1, & \text{$x \in \mathbb{R}\backslash\mathbb{Q}$} \\ 0, & \text{$x \in \mathbb{Q}$} \end{cases}$

Prove/Disprove the existence of the $\displaystyle\lim_{x\rightarrow x_0}f(x)$ while:

  1. $x_0\in \mathbb{R}\backslash\mathbb{Q}$
  2. $x_0\in \{1,-1\}$

I think the first one is no and the second one is yes but I don't know how to show it.

Please show me a formal proof if possible.

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1) Choose a sequence $(x_{n})$ in ${\bf{Q}}$ such that $x_{n}\rightarrow x_{0}$, then $f(x_{n})=0\rightarrow 0$ but $f(x_{0})=x_{0}^{2}-1\ne 0$ (it if were, then $x_{0}=1$ or $-1$, then $x_{0}$ is a rational number).

2) Say, $x_{0}=1$. Then given $\epsilon\in(0,1)$ and for all $x$ with $|x-1|<\epsilon$, if $x$ is a rational number, then $|f(x)-f(1)|=|f(x)|=0<2\epsilon$, if $x$ is an irrational number, then $|f(x)-f(1)|=|f(x)|=|x^{2}-1|=|x+1|\cdot|x-1|<2\epsilon$, in either case we have $|f(x)-f(1)|<2\epsilon$ for all $x$ with $|x-1|<\epsilon$, so $f$ is continuous at $x_{0}=1$.

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There exists a sequence of rational numbers $a_n$ approaching $x_0$. $f(a_n) = 0$ for all $n$ so the limit as $n \to \infty$ of $f(a_n)$ is $0$. Similarly there exists a sequence of irrational numbers , $b_n$, approaching $x_0$. As $n\to \infty$, we have $f(b_n) \to x_0^2 -1$. For the limit to exist at $x_0$, we need $0 = x_0^2 -1$.

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Assume $f(x)$ is continuous at $x_{0}$. Therefore, for every sequence $({x_{n}})$, we have must have $$\lim _{n\to \infty} f(x_{n}) = f(x_{0}) = x_{0}^2-1$$for every sequence $(x_{n})$ such that $x_{n} \to x_{0}$ as $n\to \infty$. But take the sequence of rational terms $x_{n}$ such that: $$x_{0}-\frac{1}{n} < x_{n} < x_{0}+\frac{1}{n}$$ Then this sequence is composed of rational terms converging to $x_{0}$, but $$\lim _{n\to \infty} f(x_{n}) = 0 \neq f(x_{0})$$

As for the second question, if you let $x$ go to $1$ or $-1$ in both expressions of $f(x)$, you will get $0$. So $f(x)$ must go to $0$.

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Let $x_0=\sqrt{2}$ and take $\{x_n\}_{n\in\mathbb{N}}\subseteq\mathbb{Q}$ such that $\lim_{n\rightarrow\infty}x_n=x_0$. Then $lim_{n\rightarrow\infty}f(x_n)=0$. Now, if we take $\{x_n\}_{n\in\mathbb{N}}\subseteq\mathbb{R}\setminus\mathbb{Q}$ such that $\lim_{n\rightarrow\infty}x_n=x_0$, then $\lim_{n\rightarrow\infty}f(x_n)=1$.

With this you can conclude that $\lim_{x\rightarrow x_0}f(x)$ does not exists if $x_0\in\mathbb{R\setminus Q}$.

Now, take $x_0=1$ and let $\{x_n\}_{n\in\mathbb{N}}$ such that $\lim_{n\rightarrow\infty}x_n=1$. You only need verify that every subsequence of $\{f(x_n)\}_{n\in\mathbb{N}}$ converge to $0$.

As $f(x_n)=0$ is constant for each $x_n\in\mathbb{Q}$ you can "avoid them". In other case you have $f(x_n)=x_n^2-1\rightarrow 0$ when $x_n\rightarrow 1$.

When $x_0=-1$, you can use a similar reasoning.

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  • $\begingroup$ Can you provide examples for such sequences? $\endgroup$ – Moshe King Dec 25 '17 at 21:08
  • $\begingroup$ For example, $(1+ 1/n) \in\mathbb{Q}$ and $\lim_{n\rightarrow\infty}(1+\frac{1}{n})=1$. But you need arbitrary sequences for a formal proof. These sequences does exists because $\mathbb{Q, R\setminus Q}$ are denses in $\mathbb{R}$. $\endgroup$ – Gödel Dec 25 '17 at 21:28
  • $\begingroup$ I meant examples for sequences that their limits is $\sqrt{2}$ $\endgroup$ – Moshe King Dec 26 '17 at 8:26
  • $\begingroup$ @GADI Yes, in this post there is a n interesting example math.stackexchange.com/questions/1021876/… $\endgroup$ – Gödel Dec 26 '17 at 15:28
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Notice that any real number $x_0$ could be approached by a sequence of rationals and a sequence of irrationals. For the rational sequence $(q_n)$ the limit of $f(q_n)$ is $0$. For the irrational sequence $(r_n)$ the limit of $f(r_n)$ is $(x_0)^{-1}$. For continuity we need the two limits match with each other, that is $x_0= 1$ or $x_0=-1$. Therefore the function is continuous at $1$ and $-1$ and discontinuous at other points.

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