Re: opponents of taylor and l'hospital ?
- From: "porky_pig_jr@xxxxxxxxxxx" <porky_pig_jr@xxxxxxxxxxx>
- Date: Mon, 5 May 2008 08:54:26 -0700 (PDT)
On May 4, 11:29 pm, lwal...@xxxxxxxxx wrote:
In this thread below, Dr. Gerry Myerson does give a valid reason
Suppose a high school senior taking the AP Calculus exam
this month sees the following problem:
lim (x->0) (sin x)/x =
A) 0
B) 1
C) -1
D) does not exist
If I were this student, I'd simply use L'Hopital's rule and find
the anwer to be (cos x)/1 = cos 0 = 1 in seconds. Why
should I care that the limit is used in the proof that
d/dx (sin x) = cos x? The question didn't ask to prove that
d/dx (sin x) = cos x -- it asked to find the limit, and the
fastest way to find the limit is L'Hopital's rule.
There are other ways to show that d/dx(sin x) = cos x, without
computing the limit of (sin x)/x first. If you start with defining
complex exponentials e^z as the series \sum_{n=0}^{|infty} z^n/n!, and
then define e^{ix} = C(x) + iS(x), it immediately follows that d/dx(S)
= C, by observation of series representations of C(x) and S(x). The
only remaining part is to show that both C and S match the regular cos
and sin we learned in high school. Once we have this in place, we can
certainly use the L'Hospital rule to show that the limit is 1.
BUT practically all Calculus textbooks derive the limit of sin(x)/x
first, using pure geometric means, and *then* use it to show that d/
dx(sin) = cos x. With this in mind, using L'Hospital rule to argue
that the limit is 1 *is* a circular reasoning. That's why you should
care how d/dx (sin x) = cos x was obtained in the first place. (Of
course there is a reason why Calculus does not do it 'series' way: you
have to have all the series related staff in place, so that's normally
done in analysis. Baby Rudin does it this way.)
In fact, to this day I don't remember how to prove the
differentiation rules for sin x and e^x -- yet I still recall
lim (x->0) (six x)/x = lim (x->0) (e^x-1)/x = 1, thanks to
L'Hopital's rule. Except for polynomial functions, I
cannot differentiate any function from first principles only.
Well, there are different ways to define e^x. Sometime you start with
ln x first, as the integral of 1/x, then define e^x as inverse
function of ln x, then all the properties including d/dx e^x = e^x
follow. Of course in case of series representation, the fact that d/dz
e^z = e^z immediately follows, by observation.
.
- References:
- opponents of taylor and l'hospital ?
- From: amy666
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