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I discovered while teaching Calc 2 that if you apply L'Hopital's rule to $\frac{x}{\sqrt{x^2+1}}$ you get $\frac{\sqrt{x^2+1}}{x}$, and if you apply L'Hopital again you get $\frac{x}{\sqrt{x^2+1}}$ back. In other words the L'Hopital operator has a cycle of order two.

EDIT (Thanks KennyTM): "I suppose the L'Hopital operator should be defined on equivalence classes of pairs $(f(x),g(x)$ of differentiable functions with the fractional equivalence: $(f(x),g(x))\equiv(h(x),k(x))$ if and only if $fk=gh$." This does not work. But it doesn't really take pairs of functions to pairs of function, either. So the first problem is to find out how it is an operator.

Has anyone ever studied this operator? Wikipedia tells me nothing.

Yes, I know it is easier to find the limit by dividing through by $x$, but some students want to apply L'Hopital to everything.

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    @George S. I meant that it has an orbit of length 2 in its action on whatever it is acting on, which at this point I am not sure of!2010-08-14

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I dunno if its been studied as a differential operator, and I kind of doubt it, but I think you could define it in a way similar to the way that you suggested.

You consider two pairs (f,g) and (h,k) equivalent if $f/g = h/k$ or rather $fk = gh$. Then the operator L acts on equivalence classes by the following operation:

$ L(f,g) = (Df, Dg) $

But for this operator to be well defined we must enforce that $L(h,k)$ is equivalent to $L(f,g)$. This means that we must have

$Df/Dg = Dh/Dk$.

This is not always true. Suppose that $(f,g) = (x, x^2)$ which is equivalent to $(h,k) = (1,x)$.

$Df/Dg = \frac{1}{2x} \neq 0 = Dh/Dk$

So the L'Hospital operator is well defined on the equivalence classes of pairs $(f,g)$ with $g$ non-constant with the equivalence relation:

$(f,g)$ equiv $(h,k) \Leftrightarrow fk = hg $ and $ Df Dk = Dh Dg $.