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

**References**:**Re: Limit question***From:*"Alan Mason" <swt@austin.rr.com>