RE: Problem with Limits; basic calculus
- To: mathgroup at smc.vnet.net
- Subject: [mg38939] RE: [mg38952] Problem with Limits; basic calculus
- From: "Florian Jaccard" <jaccardf at eicn.ch>
- Date: Wed, 22 Jan 2003 06:09:33 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Hello !
If you multiplicate your function by 1 in an intelligent way, here
E^(3x)/E^(3x), you will more easily be able to study your limits by hand.
Mathematica also appreciates the simplification before you ask the limit !
Here with Mathematica :
In[7]:=
f[x_] = Simplify[(3/E^x - E^(-3*x))/(E^(-3*x) + E^(-x))]
Out[7]=
(-1 + 3*E^(2*x))/(1 + E^(2*x))
In[8]:=
Limit[f[x], x -> Infinity]
Out[8]=
3
In[9]:=
Limit[f[x], x -> -Infinity]
Out[9]=
-1
The Plot confirms this facts !
In[5]:=
Plot[(3/E^x - E^(-3*x))/(E^(-3*x) + E^(-x)),
{x, -100, 100}]
Meilleures salutations
Florian Jaccard
professeur de Mathématiques
EICN-HES
-----Message d'origine-----
De : David Seruyange [mailto:sophtwarez at hotmail.com]
Envoyé : mar., 21. janvier 2003 13:40
À : mathgroup at smc.vnet.net
Objet : [mg38952] Problem with Limits; basic calculus
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