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Let $f(x)=|x|$ for $x\in\mathbb{Q}$ and $f(x)=1$ for $x\in\mathbb{R}/\mathbb{Q}$, where $\mathbb{Q}$ is the set of rationals. (a) Show $f$ has limits at $1$ and $-1$. Find them. (b) Show that if $c\not=1$ or $c\not=-1$, then $f$ doesn't have a limit at $c$.

Attempt:

(a) Since $f(x)$ is either $|x|$ or $1$, we have $|f(x)|\leq|x|$ $\forall x\in\mathbb{R}$. Let $\epsilon>0$, $\delta(\epsilon)>0$.

I don't know what $\delta(\epsilon)>0$ should be. But from there I think a basic outline would be something like:

Then it follows that $1<|x|<\delta$, then $|f(x)|<\epsilon$. So $\lim\limits_{x\to0}f(x)=1.

But I'm not sure.

Any help/hints would be greatly appreciated.

(b) Suppose $c\not=1$. Let $(x_n)$ be a sequence of rationals converging to $c$. Let $(y_n)$ be a sequence of irrationals converging to $c$. These sequences exist because both $\mathbb{Q}$ and $\mathbb{R}/\mathbb{Q}$ are dense in $\mathbb{R}$. Then $$\lim\limits_{n\to\infty}f(x_n)=\lim\limits_{n\to\infty}x_n=c\not=1=\lim\limits_{n\to\infty}1=\lim\limits_{n\to\infty}f(y_n).$$ By the Sequential Criterion, $f$ does not have a limit at $c\not=1$.

For $c\not=-1$, I don't know if I can use the same method.

Any help/hints would be greatly appreciated.

Desperate Fluffy
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  • It's not true that $|f(x)|\leq |x|$ for all $x\in\mathbb{R}$; consider $x=\sqrt{2}/2$. But you're on the right track. If $f(x)$ were equal to $|x|$ for all $x$, the limits would be easy to find; what you want to do is essentially to show that the 'redefinition' of $f(x)$ for irrational $x$ doesn't affect the limits at those points. – Steven Stadnicki Oct 31 '13 at 06:41
  • Let $\epsilon>0$. At x=1, choose $\delta< \min{\epsilon,1}$. For $x$ such that $|x-1|<\epsilon$, we have $0<1-\delta<x<1+\delta$ . Hence, $|f(x)-1|=0(x\in Q) or <|1+-\epsilon -1|=\epsilon$. – Worawit Tepsan Oct 31 '13 at 06:43
  • Okay; but how do I do that. I'm not very good (at all) with using the Definition of the Limit. Both limits seem straightforward, so how do I show they exist using the Definition or something equally formal? – Desperate Fluffy Oct 31 '13 at 06:45
  • For (b), you should suppose $c\neq 1,-1$ because if you suppose $c\neq 1$, $c$ can be equal $1$. – Worawit Tepsan Oct 31 '13 at 06:48
  • First, you have to guest that what is the limit value, in this case is 1. Then you have to proof that "for every $\epsilon$, there exists $\delta$ such that $|x-x_0|<\delta$ then $|f(x)-L|<\epsilon$." – Worawit Tepsan Oct 31 '13 at 06:51
  • I know the Definition of the Limit, but I have no idea how to apply it. I know I have to choose some $\epsilon$ in my proof, but it always seems so arbitrary I can't understand it. Could you elaborate? – Desperate Fluffy Oct 31 '13 at 06:54

1 Answers1

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Hint

$f(x)=-x\forall x (-\infty,0)\cap\mathbb{Q}$

$f(x)=+x \forall x [0,\infty)\cap\mathbb{Q}$

$f(x)=1\forall x\in\mathbb{Q}^c$

can you do now?

Add take $c\in\mathbb{R}$ , now say $x\to c$ via rationals then $\lim_{x\to c}f(x)=|c|$ and if $x\to c$ via irrationals then $\lim_{x\to c} f(x)=1$, for existance of limit at $c$ we need this two limit equal, so $|c|=1$ so $c=1,-1$

take $c\in\mathbb{R}\setminus\{1,-1\}$

Now use sequential argument to get a contradiction.

Myshkin
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