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Re: Re: Limit question
*To*: mathgroup at smc.vnet.net
*Subject*: [mg31537] Re: Re: Limit question
*From*: Daniel Lichtblau <danl at wolfram.com>
*Date*: Fri, 9 Nov 2001 06:13:49 -0500 (EST)
*Approved*: Steven M. Christensen <steve@smc.vnet.net>, Moderator
*References*: <9sb2du$p56$1@smc.vnet.net> <200111080957.EAA06499@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Alan Mason wrote:
>
> Hello,
> It's odd that Limit should have this hole in it-- after all,
> Limit[x^2/Exp[x], x->Infinity] = 0 is done correctly.
>
> The notebook below gives a quick and dirty routine that implements de
> l'Hopital's rule. It works correctly in the cases considered (including
> your first example) and can serve as a start. Of course, it will bomb if
> taking successive derivatives doesn't solve the problem (i.e., when de
> l'Hopital's rule is of no use).
> <snip --dl>
> For your second example, Limit[Exp[x]/Gamma[1+x], x->Infinity]=0, you'll
> need to use Sterling's formula. Limit is one of the weakest built-in
> functions of Mathematica and it needs a lot of work for the special
> functions. It can't do Limit[BesselJ[1, x]/Sqrt[x], x->Infinity] either, for
> example, although the answer is well known to be 0. Oddly, direct
> substitution BesselJ[1, x]/Sqrt[x] /. x -> Infinity does give 0. Looks like
> Limit is some kind of neglected poor relation.
> <snip --dl>
> lh := LimitHosp;
> In[63]:=
> lh[{Exp[x],x^2}, x\[Rule]Infinity]
> Out[63]=
> \[Infinity]
> In[66]:=
> lh[{Exp[x^2]/ Exp[x] ,x^3}, x\[Rule]Infinity]
> (* NOTE: {Exp[x^2], Exp[x] x^3} will bomb *)
> Out[66]=
> \[Infinity]
> In[69]:=
> lh[{x^12 , Exp[x/100]}, x\[Rule]Infinity]
> Out[69]=
> 0
> In[13]:=
> lh[{Sin[x]^3, x (1-Cos[x])}, x\[Rule]0, 0]
> Out[13]=
> 2
>
> Alan
> (**************************************************************************)
>
> "Leonard Howell" <lwhowell at knology.net> wrote in message
> news:9sb2du$p56$1 at smc.vnet.net...
> > I'm trying to evaluate Limit [Exp[x] * (1/x^2), x-> Infinity] with
> > Mathematica but can not seem to get the correct answer of Infinity. Next,
> I
> > want Limit [Exp[x] /( x!), x-> Infinity] but can't get it either. Can
> > someone please provide guidance?
> >
> > Thanks, Leonard
> >
I thought I would mention that Limit has received a bit of work. While
by no means perfect there has been some improvement in the handling of
essential singularities. The following was done in a development version
of Mathematica.
In[17]:= Limit[Exp[x]/x^2, x->Infinity]
Out[17]= Infinity
In[18]:= In[18]:= Limit[Exp[x]/(x!), x->Infinity]
Out[18]= 0
In[19]:= In[19]:= Limit[BesselJ[1,x]/Sqrt[x], x->Infinity]
Out[19]= 0
In[20]:= Limit[Exp[x^2]/(x^3*Exp[x]), x->Infinity]
Out[20]= Infinity
In[21]:= In[21]:= Limit[x^12/Exp[x/100], x->Infinity]
Out[21]= 0
In[22]:= In[22]:= Limit[Sin[x]^3/(x*(1-Cos[x])), x->0]
Out[22]= 2
The last three are handled in version 4.1 as well.
Daniel Lichtblau
Wolfram Research
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