Re: Asymptote mystery...
- To: mathgroup at smc.vnet.net
- Subject: [mg38852] Re: Asymptote mystery...
- From: atelesforos at hotmail.com (Orestis Vantzos)
- Date: Thu, 16 Jan 2003 03:19:00 -0500 (EST)
- References: <b032u1$mvn$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
For x<0, your f[x] becomes (x+2)/(-x-2) == -1
(Try FullSimplify[f[x],x<0] )
So there is no asymptote at x==-2.
In any case, a root of the denominator does not always correspond to
an vertical asymptote: if the numerator is also zero(as in this case)
then the L'Hospital rule must be applied, which might lead to a finite
value at the potential singularity.
Orestis
Mike Summers <mike at miscanthus.net> wrote in message news:<b032u1$mvn$1 at smc.vnet.net>...
> It seems to me that this:
>
> f[x] = (x + 2)/(Abs[x] - 2)
>
>
> should show asymptotes at 2 & -2.
>
> Plot[Evaluate[f[x]], {x, -5, 5}] only shows the asymptote at 2.
>
> What am I doing wrong?
>
> Thanks-- Mike