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Re: l'Hopital's Rule


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




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