Does L'Hôpital's work the other way?

With L'Hospital's rule your limit must be of the form $\dfrac 00$, so your antiderivatives must take the value $0$ at $c$. In this case you have $$\lim_{x \to c} \frac{ \int_c^x f(t) \, dt}{\int_c^x g(t) \, dt} = \lim_{x \to c} \frac{f(x)}{g(x)}$$ provided $g$ satisfies the usual hypothesis that $g(x) \not= 0$ in a deleted neighborhood of $c$.

I recently came across a situation where it was useful to go through exactly this process, so (although I'm certainly late to the party) here's an application of L'Hôpital's rule in reverse:

We have a list of distinct real numbers $\{x_0,\dots, x_n\}$. We define the $(n+1)$th nodal polynomial as $$ \omega_{n+1}(x) = (x-x_0)(x-x_1)\cdots(x-x_n) $$ Similarly, the $n$th nodal polynomial is $$ \omega_n(x) = (x-x_0)\cdots (x-x_{n-1}) $$ Now, suppose we wanted to calculate $\omega_{n+1}'(x_i)/\omega_{n}'(x_i)$ when $0 \leq i \leq n-1$. Now, we could calculate $\omega_{n}'(x_i)$ and $\omega_{n+1}'(x_i)$ explicitly and go through some tedious algebra, or we could note that because these derivatives are non-zero, we have $$ \frac{\omega_{n+1}'(x_i)}{\omega_{n}'(x_i)} = \lim_{x\to x_i} \frac{\omega_{n+1}'(x)}{\omega_{n}'(x)} = \lim_{x\to x_i} \frac{\omega_{n+1}(x)}{\omega_{n}(x)} = \lim_{x\to x_i} (x-x_{n+1}) = x_i-x_{n} $$ It is important that both $\omega_{n+1}$ and $\omega_n$ are zero at $x_i$, so that in applying L'Hôpital's rule, we intentionally produce an indeterminate form. It should be clear though that doing so allowed us to cancel factors and thus (perhaps surprisingly) saved us some work in the end.

So would this method have practical use? It certainly did for me!

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PS: If anyone is wondering, this was a handy step in proving a recursive formula involving Newton's divided differences.

No it does not, for example consider the limit $\lim_{x\to 0}\frac{\sin x}{x}$, L'Hôpital's gives you 1, but reverse L'Hôpital's goes to $-\infty$ (without introducing extra constants).