Re: Can we use L'hopital rule here?
From: Denis Feldmann (denis.feldmann_at_wanadoo.fr)
Date: 09/15/04
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Date: Wed, 15 Sep 2004 07:59:07 +0200
Bill Dubuque wrote:
> Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:
>> lihuijun <lihuijun@hotmail.com> wrote: (edited)
>>>
>>> Can we use L/hopital rule here?
>>>
>>> 0/x as x-->0
>>>
>>> 0/x^2 as x-->0
>>>
>>> If so, then the ans is 0. Please advise
>>
>> L'hopital's rule is rarely useful, but this must be
>> the most pointless application of it ever devised!
>
> Before rushing to judgment, consider the following
>
> 2 2
> sin (x) + cos (x) - 1
> lim ---------------------
> x->0 x^2
>
>
> The expression equals 0/x^2, so has the above form.
> Applying L'Hopital's rule reduces it to the trivial
> limit of 0/(2x) and thus yields the correct result
> _without_ requiring the specialized trig knowledge
> that the initial numerator is identically zero.
But then, how do you know that it is of the required form "0/0"? The same
trick would work for
2 2
sin (x) + cos (x) - 2
lim ---------------------
x->0 x^2
... except that it doesn't
> Now imagine a similar limit only with a much more
> hairy numerator that is identically zero but is not
> immediately recognizably so. As above, L'Hospital's
> rule might similarly simply reveal the correct result.
> The same remark apply to machines as well as humans.
> Computer algebra systems may encounter such limits.
>
> Further, suppose that the rule didn't apply in the
> case the numerator is identically zero. Then before
> applying the rule one would be required to verify
> also that the numerator was not identically zero.
> But the zero equivalence problem is undecidable
> in general (even so for most any nontrivial class
> of functions one encounters in calculus, see [1]).
>
> So it's a good thing the rule applies in this case.
>
> --Bill Dubuque
>
> [1] Daniel Richardson: Some Undecidable Problems Involving Elementary
> Functions of a Real Variable. J. Symb. Log. 33(4): 514-520 (1968).
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