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When I was first taught limits, we were forbidden from using L'Hopital's rule to calculate it. It was regarded as a "shortcut" and not a "real method" to find a limit. Obviously, when you have to take a limit it is often a very easy and intuitive method to find the limit (provided you check the conditions correctly).

However, this led me to think whether this is because it really is just a shortcut that is unnecessary or whether there are places where it is absolutely essential.

Of course, an example of a limit which does require L'Hopital's rule to be solved would answer my question, but especially if it's not true and you CAN always find a limit without L'Hopitals's rule, I want some kind of proof or at least understanding of this.

The Infinite One
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    L'Hopital's rule is a real method. I don't know why you say that it's not. – David Lui Mar 01 '23 at 13:56
  • I'm not saying it's not, just that it is taught that way. You are expected to solve limits without them for some reason, as if it is just a shortcut that's not "actually necessary." I don't know if this is true, which is why I asked this question – The Infinite One Mar 01 '23 at 13:58
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    It can always be done in some other way, if only just by proving the same things over and over again. But that is true for all techniques. – Thomas Andrews Mar 01 '23 at 14:07
  • Given a limit of the form $$\lim_{x\to a}\frac{f(x)}{g(x)}$$ where both $f'(x)$ and $g'(x)$ constructively exist (meaning for any $\varepsilon$ you can find $\delta$ such that $\left|\frac{f(x+h)-f(x)}{h}-f'(x)\right|<\varepsilon$ when $|h|<\delta$), and you constructively know why $f(x)\overset{x\to a}{\to}0$ (and all that for $g$ too), then you could just plug these values for your $\varepsilon$s into the $\varepsilon,\delta$ proof of l'Hopital's rule, right? And have a nice consturctive proof for your limit as your teacher probably want. – student91 Mar 01 '23 at 14:08
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    Aside: Every once in a while, you'll see a question here that is of the form: Find $\lim_{h\tp0}\frac{f(x+h)-f(x)}h,$ for some messy $f,$ and the first answer is "Use L'Hopital." – Thomas Andrews Mar 01 '23 at 14:12
  • I do not know if it is true universally, but most of the time instead of using L'Hopital's rule you can find a limit by Taylor's decomposition – Kubrick Mar 01 '23 at 14:13
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    The discussion on this question might also be of interest to you: https://math.stackexchange.com/questions/2118581/lhopitals-rule-and-frac-sin-xx?rq=1 – student91 Mar 01 '23 at 14:13
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    Any time you are using a theorem, it is just a shorthand for rewriting the proof of the theorem. So all mathematical uses of theorems are just shortcuts - something we can reference to remove redundancy. – Thomas Andrews Mar 01 '23 at 14:14
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    @Kubrick That's really what L'Hôpital's rule is; use first-order Taylor expansion. – Jakob Streipel Mar 01 '23 at 16:53
  • (adding to prets' comment) provided that we understand "the Taylor decomposition" correctly: there are several different forms of Taylor's theorem, and some of them assume greater smoothness than L'Hopital's rule. – JBL Mar 01 '23 at 16:54
  • L'Hospital's Rule is a very powerful / useful tool, but often it is misused as students don't focus on the conditions of its applicability. – Paramanand Singh Mar 04 '23 at 02:28

2 Answers2

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Regarding your first paragraph: as an instructor, I have two learning objectives related to evaluating limits.

  • Use the limit laws (sum rule, product rule, etc.) to evaluate limits of functions.
  • Use L'Hôpital's rule to evaluate limits in indeterminate form.

The first is important because we want students to know how derivatives are calculated “from scratch.” The second is important because L'Hôpital's rule is a powerful theorem.

The trouble is, when you have a very big hammer, everything looks like a nail. Students tend to bang at every limit of any quotient of functions with L'Hôpital's hammer even in situations where:

  • the limit isn't in an indeterminate form (leading to an incorrect answer)
  • other methods (for example, Taylor series or established trigonometric limits) might resolve the limit more quickly
  • using L'Hôpital's rule results in circular reasoning
  • functions of several variables are encountered, etc.

For these reasons, we want students to know when to use L'Hôpital and when not to. That doesn't diminish the rule itself, though; it just puts the rule in its proper place.

Matthew Leingang
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    This is a good answer, but it's sad that using higher-order approximations like the Taylor expansion is not on your list. – JBL Mar 01 '23 at 16:56
  • (This answer might go more broadly under the theme that some students look at mathematics as a series of formulas to be learned, whereas it's the logical structure into which the formulas fit that is what actually matters.) – JBL Mar 01 '23 at 16:59
  • @JBL In my head I included Taylor series among “other methods.” But I'll go ahead and make it explicit. – Matthew Leingang Mar 01 '23 at 18:53
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Yes, L'Hôpital is a shortcut. If you analyze why it works, in a simplified case we have the following. You want to calculate $$ \lim_{x\to0}\frac{f(x)}{g(x)}, $$ in the case where $f(0)=g(0)=0$. The Taylor expansions of $f,g$ around $0$ are $$ f(x)=f'(0)x+o(x^2),\qquad g(x)=g'(0)x+o(x^2). $$ Then $$ \frac{f(x)}{g(x)}=\frac{f'(0)x+o(x^2)}{g'(0)x+o(x^2)}=\frac{f'(0)+o(x)}{g'(0)+o(x)}\xrightarrow[x\to0]{}\frac{f'(0)}{g'(0)}. $$ As an automation system to solve limits, L'Hôpital's rule has at least two problems:

  1. It teaches zero intuition. This delves into the philosophy of how and why calculus is taught; and the sad reality is that these in many many cases calculus is taught as a set of automated tools without any understanding. At many (most?) universities students regularly pass calculus with high grades and no understanding of what a derivative or an integral is.

  2. In many applications what one needs is not the value of a limit, but an understanding of how the function is behaving at the limit. Here is an easy example of what I mean. Consider $$ \lim_{x\to0}\frac{\sin x}x. $$ In many calculus courses this is an automatic L'Hôpital limit, the quotient of the derivatives is $\frac{\cos x}1=\cos x$, and $\cos 0=1$ (that the students will probably obtain with their calculator). If instead one uses approximations, the limit can be seen as $$ \frac{\sin x}x=\frac{x-\frac{x^3}6+o(x^5)}x=1-\frac{x^2}6+o(x^4), $$ and we learn that close to $x=0$ the function behaves as $1-\frac{x^2}6$ (telling us not only that the limit is $1$ but, among other things, that the approximation is quadratic).

Martin Argerami
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  • Here is an even simpler case: if $f$ and $g$ are linear functions with $f(a) = g(a) = 0$, then $\lim_{x\to a} \frac{f(x)}{g(x)}$ is the quotient of the slopes of $f$ and $g$. This can be shown with algebra. ¶ In general, we replace “linear” with “differentiable” and “slopes” with “derivatives.” In the limit it all works out, thanks to the Mean Value Theorem. – Matthew Leingang Mar 01 '23 at 19:56
  • This is not exactly the L'Hospital's Rule, but rather a much simpler variant of it. The rule talks about limit of $f'(x) /g'(x) $ and not about $f'(0)/g'(0)$. The version you are talking of is handled just by using definition of derivative – Paramanand Singh Mar 04 '23 at 02:25
  • @ParamanandSingh: I'm aware, which is why I wrote "in a simplified version". Adding more notation to cover the more general cases would only make the argument longer while not changing the main point to be main, which is that L'Hôpital lets you calculate limits blind, without seeing what happens. – Martin Argerami Mar 04 '23 at 02:28
  • Sorry to say but I did miss the phrase "in a simplified version". If you wish I can delete my comment. – Paramanand Singh Mar 04 '23 at 02:29