Is L'Hopital's rule just a shortcut? OR Are there limits that CANNOT be done without using L'Hopital's rule?

Question

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.

Answer 1

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).

Answer 2

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.