Re: Can we use L'hopital rule here?

From: The World Wide Wade (waderameyxiii_at_comcast.remove13.net)
Date: 09/16/04 

Date: Thu, 16 Sep 2004 14:37:53 -0700

In article <a8f62c00.0409151446.175828fa@posting.google.com>,
 night@fas.harvard.edu (Christopher) wrote:

> 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?

I doubt one "knows" the values of sin and cos at 0, and that these are
continuous functions, without knowing sin^2 + cos^2 = 1. But yes, you could
remember those facts in isolation.

> > 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.

True, and as I wrote to Bill, I shouldn't have gotten involved with that
issue, because the only point I was trying to make is that it is silly to
use LHR to evaluate lim_s->0 0/s^2. Which it is.

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