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Re: Evaluation of limit

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  • Subject: [mg115746] Re: Evaluation of limit
  • From: Bob Hanlon <hanlonr at>
  • Date: Wed, 19 Jan 2011 05:31:28 -0500 (EST)


"8.0 for Mac OS X x86 (64-bit) (November 6, 2010)"

Limit[Cosh[x] + Sinh[x]^2*Log[Tanh[x/2]],
 x -> Infinity]


Bob Hanlon

---- "Dr. C. S. Jog" <jogc at> wrote: 

If I give the command

In[1]:= Limit[Cosh[x]+Sinh[x]^2*Log[Tanh[x/2]],x->Infinity]

I get
                                 x          2
Out[1]= Limit[Cosh[x] + Log[Tanh[-]] Sinh[x] , x -> Infinity]
instead of zero. This result can be verified numerically (of course, by 
using higher precision than the default).


C. S. Jog

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