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Re: l'Hopital's Rule



"Kevin J. McCann" wrote:
> 
> If you divide numerator and denominator by a, Mathematica will then give the
> correct answer of 2a. Also if a is replaced by a number, the correct
> answer is given. I would call this a bug in the Limit.
> 
> Kevin
> 
> 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

Interestingly, Macsyma does the same thing. I should
add this to my list of system-independent bugs.


Here's my explanation:
  The numerator evaluates to 0.

The denominator does not evaluate to zero, since a-sqrt(a*a) cannot be
(correctly) automatically simplified. Indeed, it may be either 0 or 2*a,
depending
on the choice of sign of the square root. Choosing either particular
answer can be used to produce any number of falsehoods.
Curiously, here, NOT making a choice leads to problems, too!

The correct limit is not 2a, but the PAIR OF POSSIBLE
ANSWERS {0 , 2a}, depending on the choice of root.

One thing you can do (in macsyma... there is something
similar in Mathematica, but I forget its name.)
radcan(%)  gives sqrt(a)*sqrt(x)+a.  Radcan makes a
choice (sometimes not the one you want) for signs of radicals.

Given that expression, limit produces 2*a.


My evaluation of this:  it is more than a bug in limit:
it is a bug in the concept of simplification in computer
algebra systems... one based on the misconception that
all algebraic expressions are single-valued...
which causes limit's test for the zero-ness of the denominator
to fail. 

Also, I should add, if you think there is only one answer here,
then YOU are committing a bug, too.

RJF


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