calculus - Does L'Hôpital's work the other way?

Hello fellows,

As referred in Wikipedia (see the specified criteria there), L'Hôpital's rule says,

$$
\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}
$$

As

$$
\lim_{x\to c}\frac{f'(x)}{g'(x)}=
\lim_{x\to c}\frac{\int f'(x)\ dx}{\int g'(x)\ dx}
$$

Just out of curiosity, can you integrate instead of taking a derivative?
Does

$$

\lim_{x\to c}\frac{f(x)}{g(x)}=
\lim_{x\to c}\frac{\int f(x)\ dx}{\int g(x)\ dx}
$$

work? (given the specifications in Wikipedia only the other way around: the function must be integrable by some method, etc.) When? Would it have any practical use? I hope this doesn't sound stupid, it just occurred to me, and I can't find the answer myself.

Edit

(In response to the comments and answers.)

Take 2 functions $f$ and $g$. When is

$$
\lim_{x\to c}\frac{f(x)}{g(x)}=
\lim_{x\to c}\frac{\int_x^c f(a)\ da}{\int_x^c g(a)\ da}
$$

true?

Not saying that it always works, however, it sometimes may help. Sometimes one can apply l'Hôpital's even when an indefinite form isn't reached. Maybe this only works on exceptional cases.

Most functions are simplified by taking their derivative, but it may happen by integration as well (say $\int \frac1{x^2}\ dx=-\frac1x+C$, that is simpler). In a few of those cases, integrating functions of both nominator and denominator may simplify.

What do those (hypothetical) functions have to make it work? And even in those cases, is is ever useful? How? Why/why not?

Answer

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!

---

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