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Re: Problem with Limits; basic calculus

  • To: mathgroup at smc.vnet.net
  • Subject: [mg38955] Re: Problem with Limits; basic calculus
  • From: atelesforos at hotmail.com (Orestis Vantzos)
  • Date: Wed, 22 Jan 2003 06:11:22 -0500 (EST)
  • References: <b0jfs1$t8q$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Since Exp[-x] goes to zero as x goes to +infinity, the dominant term
is not Exp[-3x]==Exp[-x]^3 but Exp[-x], so 3 is the real value of the
limit.

Using Simplify on the original expression yields:
-1 + 3 Exp[2x]
---------------
1+ 2 Exp[2x]
It is clear (use L'Hospital rule for formal proof) that the limit is
indeed 3.

Finally, by applying FullSimplify we get 1 + 2Tanh[x] and since
Limit[Tanh[x],x->inf]==1 we also get 3.

Orestis


sophtwarez at hotmail.com (David Seruyange) wrote in message news:<b0jfs1$t8q$1 at smc.vnet.net>...
> Hey all - I'm taking a basic calculus course that uses Mathematica. 
> We have been studying limits and I have been using the Limit function
> to check if my answers are correct.
> We were given the following function and asked to determine a limit:
> (3E^(-x) - E^(-3x)) / (E^(-3x) + E^(-x))
> 
> Usually the approach is to select the dominant terms, factor and then
> determine the limit.  My initial reason had me select -E^(-3x) in the
> numerator and E^(-3x) in the denominator.  Factoring the terms would
> yield -1, thus the limit for x->infinity.  But I plotted the function
> and the real answer is somewhere near 3.
> I then tried to use the Limit function which is not producing an
> answer (perhaps I'm not sure of the usage).
> 
> Any help is greatly appreciated, 
> 
> David Seruyange


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