Teacher2Teacher |
Q&A #6003 |
From: Joshua
(for Teacher2Teacher Service)
Date: Apr 02, 2001 at 16:49:13
Subject: Re: 2nd semester Calculus
Hi Kerry, When I teach L'Hopital's rule, I try to base it on things that they've already learned about local linear approximations and/or Taylor series. For example, why is the limit x->0 of (sin x)/x equal to zero? Well, we used that fact already to prove that the derivative of sin x is cos x, so it's really circular reasoning to use the derivative of sin x in L'Hopital's rule to prove this limit, but it will illustrate the idea anyway. Since the slope of sin x at the origin is 1, to first order in x, sin x = x near the origin. Since x is small, any x^2 term will be negligible by comparison. So, sin x = (derivative of sin x)*x, and hence sin x / x = (derivative of sin x) = 1. That illustrates the principle in general: if f(a) = 0, then near a, f(x) is approximately equal to 0 + f'(a) * (x-a). That's the equation of the tangent line to f(x) at the point a. That's also first step in developing Taylor/MacLaurin series! If g(a) = 0, g(x) is approximately equal to its tangent line, 0 + g'(a) * (x-a). Dividing those two gives f'(a) / g'(a), which is just L'Hopital's rule. In practice, in my physics courses people rarely used L'Hopital's rule by name. Instead, they used the idea: They took the first nonzero term of the Taylor series of both top and bottom and then divided those to find the limit. I'm having a hard time thinking of a specific example of that offhand, though. I hope that helps! Have fun teaching it, and feel free to write back if you'd like to continue this conversation. -Joshua, for the T2T service
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