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I'm preparing a design/teaching experiment for my Curriculum Design Course right now. I've decided to cover L'Hopital's rule with a student I've been working with for a year, so skill wise he's ready to go. And I'm probably only going to focus on the form 0/0 in my tasks.

I'm trying to utilize an APOS & RME framework to guide my tasks - but I'm having an issue coming up with the necessity for L'Hopital's rule. I know we need it to completely evaluate indeterminate forms - but I'm sure my student will just be fine leaving it and saying he's done.

Any insight would be appreciated, and if I'm posting this is the wrong spot or if there's a better place to go I would like to know.

Thanks!

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    $\begingroup$ In a workshop toolbox there are many wrenches, pliers, screwdrivers etc. that aren't used every day, but the day you need one of them and you don't find it, you'll have to spend a lot of time to go to the shop, find what you need and go back to what you were doing. That's it ;-) $\endgroup$
    – Massimo Ortolano
    Nov 8, 2017 at 21:42
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    $\begingroup$ What are "APOS & RME"? $\endgroup$
    – Nick C
    Nov 9, 2017 at 1:36
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    $\begingroup$ It's not really needed, I think, except by students whose skills in, or patience for, analysis aren't very strong. It can make very complicated expressions easy to deal with, but very complicated expressions are not that common. For calculus students, it makes otherwise difficult limits easy to evaluate. These include important ones, such as the ratios of exponential, logarithmic and power/polynomial functions. However, it is taught to students as an easy procedure to uses in all applicable cases that serves as a convenient substitute for understanding & thinking. $\endgroup$
    – Raciquel
    Nov 9, 2017 at 2:58
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    $\begingroup$ Possibly related: matheducators.stackexchange.com/questions/7274/… $\endgroup$
    – Aeryk
    Nov 9, 2017 at 16:50
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    $\begingroup$ Well, in the French curriculum L'Hopital's Rule is not taught at all, so I will have a hard time explaining its necessity. We mostly use Taylor series in case you would apply it, I guess. $\endgroup$
    – Benoît Kloeckner
    Nov 9, 2017 at 21:45

3 Answers 3

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This is an answer to the title. Defining APOS & RME framework would make answering the question easier.

As Massimo Ortolano mentioned in a comment, l'Hôpital's rule is one tool in a box. Maybe you use it often, maybe rarely, but it is very nice when you can use it.

Just the other day, when trying to understand how ultrasound mediated electrical impedance tomography works, I invoked l'Hôpital. Certainly I could have derived the proper limit in another way, but l'H is easy to remember and intuitive and easy to use.

As a bonus my colleague (with an engineering background) knew the rule and so could follow the calculation with fair ease. Generalizing, l'H is good to know because many people in technical fields know it, so it makes communication smoother.

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When teaching calculus I like to include L'Hospital's Rule because it exemplifies how the subject is an impressive arsenal of calculational methods, and because it is easy to explain why this rule is not a arbitrary trick.

The form $\displaystyle{\frac{f'(a)}{g'(a)}}$ looks magical on the surface, but when viewed as $\displaystyle{\frac{f(a)+f'(a)(x-a)+O(x-a)^2}{g(a)+g'(a)(x-a)+O(x-a)^2}}$ where $f(a)=g(a)=0$, one understands how it is a simple application of the calculus concept "linearization."

The efficacy of calculus as a subject is that it is a toolbox of shortcuts that streamline calculations. L'Hospital is a lesson that focuses on one particular tool that is widely used, difficult to forget, yet is easily grounded in the concept of linearization.

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If you'd like to graph y=x*lnx, it would be useful to know where the function is headed as x->0 (from the right). L'H can be used here, though not the 0/0 form.

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    $\begingroup$ If you're willing to accept that the limit doesn't oscillate (i.e. the limit exists finitely or signed-infinitely), it's easy to show students what the limit is by using the sequence $e^{-1},$ $e^{-2},$ $e^{-3},$ $\ldots,$ since the outputs for these values are $- \frac{1}{e},$ $- \frac{2}{e^2},$ $- \frac{3}{e^3},$ $\ldots,$ a sequence that clearly approaches $0$ from below the $x$-axis. (This is intended as a useful teaching idea, not something that nails down a rigorous proof.) $\endgroup$
    – Dave L Renfro
    Nov 15, 2017 at 1:24

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