[Date Index]
[Thread Index]
[Author Index]
Re: l'Hopital's Rule
*To*: mathgroup at smc.vnet.net
*Subject*: [mg24336] Re: [mg24309] l'Hopital's Rule
*From*: Rob Pratt <rpratt at email.unc.edu>
*Date*: Sun, 9 Jul 2000 04:52:43 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
First, we must assume that a != 0 since otherwise the function of x is
undefined. Also, we need not employ L'Hopital's Rule if instead we
multiply top and bottom by the conjugate of the denominator, namely
a + Sqrt[a*x].
Limit[(a^2 - a*x)/(a - Sqrt[a*x]), x -> a]
Limit[(a^2 - a*x)/(a - Sqrt[a*x]) (a + Sqrt[a*x])/(a + Sqrt[a*x]), x -> a]
Limit[(a^2 - a*x)(a + Sqrt[a*x])/(a^2 - a*x), x -> a]
Limit[a + Sqrt[a*x], x -> a] (we used a != 0 here too)
a + Sqrt[a^2] = a + Abs[a]
If a > 0, we get 2a. If a < 0, we get 0.
Rob Pratt
Department of Operations Research
The University of North Carolina at Chapel Hill
rpratt at email.unc.edu
http://www.unc.edu/~rpratt/
On Fri, 7 Jul 2000 heathw at in-tch.com wrote:
> Hi,
> When I input this:
> Limit[(a^2 - a*x)/(a - Sqrt[a*x]), x -> a]
> The output is:
> 0
> The output should be 2*a.
> Can't Mathematica 4 use l'Hopital's Rule?
> Thanks,
> Heath
Prev by Date:
**Re: PrimeQ queries**
Next by Date:
**Re: combining graphics**
Previous by thread:
**Re: l'Hopital's Rule**
Next by thread:
**Re: l'Hopital's Rule**
| |