# Precise definition of a limit

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• July 2nd 2009, 04:42 PM
Chokfull
Precise definition of a limit
I need help with this problem...
prove that the limit

$lim x-->0 f(x)$
such that $f(x) = 0$ when $x$ is rational; $1$ when $x$ is irrational

does not exist. Thanks to anyone who helps!

yes, I think I should learn LaTex...
(Nerd)
• July 2nd 2009, 04:59 PM
Plato
You have some real issues with notation.
What you have posted is almost unreadable.
If you want to receive real help here then you should learn LaTex.

Your function is $
f(x) = \left\{ {\begin{array}{*{20}c}
{0,} & {x \in \mathbb{Q}} \\
{1,} & {else} \\

\end{array} } \right.$
.

Now you need to recall that any neighborhood of 0 contains infinitely many of both rational and irrational numbers.
There has got to a contradiction there if that limit exists.
• July 2nd 2009, 07:28 PM
Bruno J.
This is easy with sequences.

Show that there is a sequence $(x_i)_{i\in \mathbb{N}}$ of irrational points converging to 0; then if $f$ is continuous at 0 we must have $\lim f(x_i) = f(0) = 0$; but $\lim f(x_i) = \lim 1 = 1$ hence $f$ is not continuous at 0 (or anywhere for that matter).

As a side note, a funny thing is that the function

$g(x) = \left\{ {\begin{array}{*{20}c}
{0,} & {x \in \mathbb{Q}} \\
{x,} & {else} \\

\end{array} } \right.$

is continuous at 0.

I also second Plato's advice to learn LaTeX. (At least you're trying a bit!)
• July 2nd 2009, 09:42 PM
matheagle
Both sets are dense.
So any interval about zero, no matter how small, will always contain rationals and irrational numbers.
Thus for all x in $(-\epsilon, \epsilon)$ f(x) will be both 0 and 1.