How many times do you have to use L'Hôpital's rule?
Sure. Do you want $18$ times? Then consider the limit$$\lim_{x\to0}\frac{x^{18}}{x^{18}}$$or the non-trivial example$$\lim_{x\to0}\frac{\sin(x^{18})}{1-\cos(x^9)}.$$For the case in which you always get $\frac00$, consider the function$$\begin{array}{rccc}f\colon&\mathbb{R}&\longrightarrow&\mathbb{R}\\&x&\mapsto&\begin{cases}e^{-1/x^2}&\text{ if }x\neq0\\0&\text{ if }x=0\end{cases}\end{array}$$and the limit$$\lim_{x\to0}\frac{f(x)}{f(x)}$$or the non-trivial example$$\lim_{x\to0}\frac{f(x)}{f(x^2)}.$$
A couple of rather famous limits that each require 7 applications of L’Hôpital’s rule (unless evaluated by another method) are
$$ \lim_{x \rightarrow 0} \,\frac{\tan{(\sin x)} \; - \; \sin{(\tan x)}}{x^7} \;\;\; \text{and} \;\;\; \lim_{x \rightarrow 0} \, \frac{\tan{(\sin x)} \; - \; \sin{(\tan x)}}{\arctan{(\arcsin x)} \; - \; \arcsin{(\arctan x)}} \;\; $$
These two limits are discussed in the chronologically listed references below, with [11] being a generalization of the tan/sin and arctan/arcsin version. (Both [10] and [11] were brought to my attention by user21820.) Another limit that requires 7 applications of L’Hôpital’s rule is the following, which I mentioned (in an incorrect way, however) at the end of [6]:
$$ \lim_{x \rightarrow 0} \,\frac{\tan x \; – \; 24\tan \frac{x}{2} \; - 4\sin x \; + \; 15x}{x^7} $$
[1] sci.math, 13 February 2000
[2] sci.math, 16 April 2000
[3] sci.math, 11 July 2000
[4] sci.math, 13 August 2001
[5] sci.math, 12 February 2005
[6] sci.math, 27 December 2007
[7] sci.math, 7 October 2008
[8] A question regarding a claim of V. I. Arnold, mathoverflow, 8 April 2010.
[9] How find this limit $\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$, Mathematics Stack Exchange, 2 November 2013.
[10] Limit of $\dfrac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))}$ as $x \rightarrow 0$, Mathematics Stack Exchange, 26 May 2014.
[11] $\lim_{x \to 0} \dfrac{f(x)-g(x)}{g^{-1}(x)-f^{-1}(x)} = 1$ for any $f,g \in C^1$ that are tangent to $\text{id}$ at $0$ with some simple condition, Mathematics Stack Exchange, 26 May 2014.
To me, the simplest (nontrivial) way to do this is to exploit functions' representations as power series. For instance, begin with:
$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \dots = \sum_{i=0}^{\infty} \frac{x^i}{i!}.$$
To cook up an interesting L'Hospital problem, subtract off the first few terms of this series expansion from $e^x$ and divide by an appropriate term. All the following are classical Calculus I examples which are inspired by the above series expansion: \begin{align*}\lim_{x \to 0} \frac{e^x - 1}{x} &\qquad \text{(requires 1 use of L'H)} \\ \lim_{x \to 0} \frac{e^x - 1- x}{x^2} &\qquad \text{(requires 2 uses of L'H)} \\ \lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3} &\qquad \text{(requires 3 uses of L'H)} \end{align*} and so forth. You can pick any function you like in place of $e^x$, of course, so long as it has enough derivatives to play with.
You can also use this approach to cook up slightly more interesting examples. For instance, we could subtract off the appropriate terms from $e^x$ and $\cos(x)$ to get their series expansions to be $Cx^2 + [\text{lower order terms}]$. Specifically, $$\lim_{x \to 0} \frac{e^x - 1 - x}{\cos(x)- 1 }$$ has a nonzero limit and requires two uses of L'Hospital's rule. If we wanted four, we could have subtracted out the $x^2$ and $x^3$ terms from the $e^x$ expansion and the $x^2$ term from the $\cos(x)$ expansion.
What I like about this approach:
- The examples are nontrivial, in the sense that no elementary algebraic techniques will save you from having to use L'Hospital's rule.
- You can immediately tell how many uses of L'Hospital's rule will be required.
- I think it conveys something important both about Taylor series representations of functions and about how L'Hospital's rule works.