Precise definition of a limit

**Chokfull** · July 2nd 2009, 04:42 PM

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...

**Plato** · July 2nd 2009, 04:59 PM

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.

**Bruno J.** · July 2nd 2009, 07:28 PM

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!)

**matheagle** · July 2nd 2009, 09:42 PM

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.

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