Re: Floor function

On Mar 30, 12:07 pm, Virgil <Vir...@xxxxxxxxx> wrote:
In article
<7e99c484-01b2-4906-9768-820aef532...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,







conrad <con...@xxxxxxxxxx> wrote:
On Mar 30, 9:42 am, José Carlos Santos <jcsan...@xxxxxxxx> wrote:
On 30-03-2008 15:15, conrad wrote:

My book says: f(x) = floor(x) + floor(-x)
and asks for what values of 'a' does lim{x->a} f(x) exist?

If x is an integer then: f(x) = 0

But if x is any real number that is not an integer, then:
f(x) = -1

Then it says, the lim{x->a} f(x) exists and is equal to -1 for
/all values/ of a.

/all values/ to me signals the integers too. But if f is
discontinuous at all integers, how can this be?

What's the problem? Yes, your function is discontinuous at all
integers. And... ? Do you see any contradiction? If you do, which
contradiction are you talking about?

If f(x) = 0 where 'a' is some integer
but f(x) = -1 where 'a' is some real that is not an integer
then how can lim{x->a} f(x) be -1 for all values of 'a'?

A limit as x -> a specifically ignores x = a, and only considers values
of x NEAR a.


I thought the epsilon/delta def'n of limit was (for example) lim x->1
of f(x) = r <==>
for all epsilon > 0, there exists delta > 0 such that
|(1-x)| < delta ==> |(f(x)-r| < epsilon.

Taking epsilon = 1/2, then no such delta exists, since no matter how
small delta
is, f(1) = 0 and f(1+[delta/2]) = -1. Consequently, the lim x-> of
f(x) is not def'd at 1.

What am I missing?
.