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**

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:

$\displaystyle\lim_{x \to \infty}\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}} = \lim_{x \to \infty}\dfrac{1-e^{-2x}}{1+e^{-2x}} = \dfrac{1+0}{1+0} = 1$.

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:

$\displaystyle\lim_{x \to 0^+}x \ln x = \lim_{x \to 0^+}\dfrac{\ln x}{1/x} = \lim_{x \to 0^+}\dfrac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0$.

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.

**Attribution***Source : Link , Question Author : Bombyx mori , Answer Author : JimmyK4542*