When does L' Hopital's rule fail?

Question

This thought jumped out of me during my calculus teaching seminar.

It is well known that the classical L'Hospital rule claims that for the $\frac{0}{0}$ indeterminate case, we have: $$ \lim_{x\rightarrow A}\frac{f(x)}{g(x)}=\lim_{x\rightarrow A}\frac{f'(x)}{g'(x)} $$ where the later could take any value including $\infty$. Here we assume that right hand side limit exist.

However, to apply it one often has to take the derivative of $f'(x)$ again at $A$, and in principle one assumes by repeatedly applying this rule we can resolve the problem by plug in the value into the function's derivative at $A$. My question is, what if the student ask if it is possible for $\lim_{x\rightarrow A} f(x),\lim_{x\rightarrow A} f'(x)\cdots \lim^{n}_{x\rightarrow A}f^{n}(x)$ be all zero for any $n$, so the rule 'fails'. How should we answer the question properly?

For example, consider the well-known non-analytic smooth function: $$f(x)= \begin{cases} e^{-1/x}& x> 0\\ 0& x\le 0 \end{cases} $$ It is a trivial exercise to verify that $f^{n}(0)=0$ for any $n\in \mathbb{N}$. Now using L'Hospital rule we compute (as if we are a low level student) $$ 1=\lim_{x\rightarrow 0^{+}}\frac{f(x)}{f(x)}=\lim_{x\rightarrow 0^{+}}\frac{f'(x)}{f'(x)}=\lim_{x\rightarrow 0^{+}}\frac{f''(x)}{f''(x)}\cdots =\frac{0}{0}=? $$ as the chain does not stop if the student applies the rule faithfully and blindly. This is a silly example, but in general for non-analytical functions I think this kind of thing could happen. And there should be more non-analytical functions than analytical functions. Is there a way for us to resolve this at introductory calculus level, so that the student know what to do, without introducing `confusing concepts' like $\epsilon-\delta$ language, Cauchy mean value theorem, Taylor series, and infinitesimals?

Answer 1

Even for analytical functions, this kind of thing can happen.

Consider $\displaystyle\lim_{x \to \infty}\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$.

Blindly applying L'Hopital's Rule repeatedly gives: $\displaystyle\lim_{x \to \infty}\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}} = \lim_{x \to \infty}\dfrac{e^{x}+e^{-x}}{e^{x}-e^{-x}} = \lim_{x \to \infty}\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}} = \cdots$.

But if we divide the numerator and denominator by $e^x$ we get: $\displaystyle\lim_{x \to \infty}\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}} = \lim_{x \to \infty}\dfrac{1-e^{-2x}}{1+e^{-2x}} = \dfrac{1+0}{1+0} = 1$.

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Also, consider $\displaystyle\lim_{x \to 0^+}x \ln x = \lim_{x \to 0^+}\dfrac{\ln x}{1/x}$.

Blindly applying L'Hopital's Rule repeatedly gives: $\displaystyle\lim_{x \to 0^+}x \ln x = \lim_{x \to 0^+}\dfrac{\ln x}{1/x} = \lim_{x \to 0^+}\dfrac{1/x}{-1/x^2} = \lim_{x \to 0^+}\dfrac{-1/x^2}{2/x^3} = \lim_{x \to 0^+}\dfrac{2/x^3}{-6/x^4} = \cdots$.

But it we stop after applying L'Hopital's Rule once and simplify stuff, we get: $\displaystyle\lim_{x \to 0^+}x \ln x = \lim_{x \to 0^+}\dfrac{\ln x}{1/x} = \lim_{x \to 0^+}\dfrac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0$.

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In both of these problems, the solution was to use basic algebra instead of just L'Hopital's Rule. The techniques taught in introductory calculus will not solve every limit problem in the world, but they will solve the problems encountered in introductory calculus. The important thing for students is to know many techniques and learn to figure out which ones will work for a given problem. Many students learn L'Hopital's Rule and then forget how to use every other tool. This is why after teaching L'Hopital's Rule, you should throw in a few examples where L'Hopital's Rule fails. This way, they think of L'Hopital's Rule as just another tool instead of magic.

Answer 2

A priori l'Hospital's rule is not meant to be a magic rule to solve limits ; you can write every limit in the universe as a ratio of two functions, and if these two functions happen to be infinitely-differentiable-non-analytic functions like yours, then applying l'Hospital's rule won't be much of use.

At the introductory calculus level, what is most important for students is not to learn all those rules by heart and throw them on paper (which is useful to get grades, but completely useless outside the classroom for most of them), but rather to use their brains and figure out a solution. I don't think that writing $$ \lim_{x \to 0} \frac{e^{-\frac 1x}}{e^{-\frac 1{x^2}}} $$ by writing it as $e^{\frac 1{x^2} - \frac 1x}$... okay I had a comment about this but it kind of went wrong when I found the trick.

And yes, there will always be issues with such functions ; it could be a good exercise to come up with functions whose limit might exist but be hard to compute.

Hope that helps,