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Re: Limit[f[x], x->a] vs. f[a]. When are they equal?

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  • Subject: [mg118374] Re: Limit[f[x], x->a] vs. f[a]. When are they equal?
  • From: Richard Fateman <fateman at eecs.berkeley.edu>
  • Date: Mon, 25 Apr 2011 07:27:27 -0400 (EDT)

On 4/24/2011 4:57 PM, Richard Fateman wrote:
> ...

>
> First, a bug?
>
> Series[1/(1-Exp[I x]), {x, 0, 4}]   returns unevaluated. A bug?

I tried this again.  Somehow I managed to insert the denominator as a 
TextCell, and didn't notice it.
Maybe I cut/pasted from the wrong place.  Anyway, I checked and saw the
  InputForm
TextCell["(1-Exp[I x])"]^(-1)

Sorry.


<snip>


But now try this:

s= 1/(1-Exp[I x])

Limit[s,x->0]  gives ComplexInfinity

but

Limit [Series[s,{x,0,4}],x->0  gives  I * Infinity.

Can these both be right?

but then, Sin[2*x] = 2*Cos[x]*Sin[x],  ... yet Limit[ ..,x->Infinity]   
gives 2 different  answers. oh well.




RJF


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