Re: Can we use L'hopital rule here?

From: Christopher (night_at_fas.harvard.edu)
Date: 09/15/04 

Date: 15 Sep 2004 15:46:44 -0700

The World Wide Wade wrote ...
> In article <y8zd60oeaxh.fsf@nestle.csail.mit.edu>,
> Bill Dubuque <wgd@nestle.csail.mit.edu> wrote:
>
> > 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.
>
> That's dubious for two reasons. First, to apply LHR, you must verify the
> numerator -> 0; how will you do that without knowing sin^2 + cos^2 = 1?

That's easy. You know sin and cos are continuous and you know their
values at 0. What kind of calculus did you teach, anyway?

> Second, we're to believe someone has forgotten the most basic of trig
> identities, sort of like forgetting your own name, and yet knows how to
> apply the chain rule and remember the derivatives of sin and cos correctly?

He was obviously just using a simple example to make the point. See
below if you insist on a more complicated one.

> > 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.
>
> You must verify the hairy numerator -> 0 to use LHR. If it's identically 0
> but is unrecognizable as such, how will you know it -> 0?

Consider instead:

(4(cos(x)-cos(3x))/(1+sec(x)) - 8((1-cos(x))/sec(x))^2)^2 -
(1-cos(4x))^2

It takes about 10 seconds to verify that it's continuous at 0, and has
a value of 0 there. Now, I don't know if taking the derivative of this
expression and evaluating it at x=0 is easier than showing it's
identically equal to 0, but I would believe someone if they told me it
was easier for them.