Re: Another naive question - limts are meant for not crossing or not reaching?
- From: David W. Cantrell <DWCantrell@xxxxxxxxxxx>
- Date: 25 Nov 2007 13:53:29 GMT
Venkat Reddy <vreddyp@xxxxxxxxx> wrote:
On Nov 25, 10:29 am, quasi <qu...@xxxxxxxx> wrote:
On Sat, 24 Nov 2007 21:22:11 -0800 (PST), Venkat Reddy
<vred...@xxxxxxxxx> wrote:
On Nov 25, 9:03 am, William Hughes <wpihug...@xxxxxxxxxxx> wrote:
On Nov 24, 10:37 pm, Venkat Reddy <vred...@xxxxxxxxx> wrote:
What do we mean by lim x->oo 1/2^x = 0 ? That It never crosses
zero or it never reaches zero?
Fo me, if it can reach then it can cross as well, for there is no
specific change at that point which prevents crossing. This
indicates it never reaches zero. So the equality must be wrong.
The sequence 1/2,1/4,1/8... never reaches zero.
Desite this fact, the equality is not wrong.
The lim x->oo 1/2^x does not mean "the value that
the sequence 1/2^x reaches" it is a real number L
with the property that the sequence
1/2,1/4,1/8,... gets and stays arbitrarily close to L
(i.e. the difference between L and the series
can be shown to be less than any number greater
than zero). It is not hard to show that the sequence gets
and stays arbitrarily close to 0, and that the sequence
does not get and stay arbitrarily close to any
other real number. The conclusion is that L=0.
Ok, the equality is not conclusive about x but about some other number
called L to which x is getting and staying arbitrarily close.
Right.
There's hope for you yet.
Then, if the "limit" has nothng to say about whether x reaches zero or
not, I guess we both agree that x never reaches zero. Is that right?
In this case, yes.
But as has been pointed out, even if a sequence (some other sequence)
does "hit zero" or "crosses zero", the limit would still be zero, as
long as the sequence gets and stays close to zero -- arbitrarily
close.
Got it. Some functions could be taking values on both side of zero,
but converging on zero as a limt. I think I've got the answer in my
case - it never hits zero. Thanks.
Whether 1/2^x actually "hits zero" or not depends on what is used for the
domain of the function. But let's look at another example first:
Let f(x) = Sqrt(1 - x^2) with domain (-1, 1). Then the limit of f(x) as x
approaches 1 from the left is 0, but f never hits 0 since it is strictly
positive on its domain. But instead, if we were to use the closed interval
[-1, 1] for the domain of f, then f would hit zero when x = 1.
Now for your function g(x) = 1/2^x, the situation is similar. If we take
its domain to be just the real numbers, (-oo, +oo), then the limit of g(x)
as x approaches +oo is 0, but g never hits 0 since it is strictly positive
on its domain. But instead, if we were to use the extended reals,
[-oo, +oo], for the domain of g, then g would hit zero when x = +oo,
that is, g(+oo) = 0.
David
.
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