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.

  • $\begingroup$ Well, a Taylor expansion of the denominator of your original function about a point c would go like √(1+c²)+c(x-c)/√(1+c²)+… so that's one source of your observed behavior. $\endgroup$ – J. M. is a poor mathematician Aug 14 '10 at 14:20
  • 4
    $\begingroup$ I wonder if this operator can be well-defined, e.g. $(x, x^2) \mapsto (1, 2x)$ but $(x^2, x^3) \mapsto (2x, 3x^2) \equiv (2, 3x)$ but clearly $(1, 2x) \not\equiv (2, 3x)$. $\endgroup$ – kennytm Aug 14 '10 at 14:37
  • $\begingroup$ @KennyTM You are right. I edited the question to reflect this. $\endgroup$ – SixWingedSeraph Aug 14 '10 at 15:28
  • 3
    $\begingroup$ What do you mean by "In other words the L'Hopital operator has a cycle of order two."?? That its square is the identity operation? If you apply L'Hopital twice to sinxx\frac{sin x}{x}, you do not get back the same thing. There are examples when it needs to be applied a number of times to get a simple expression. So the operator is not periodic. And again, if you want to define the operator formally, there is a pre-condition that both the numerator and the denomiator should vanish at the point. $\endgroup$ – user1119 Aug 14 '10 at 15:44
  • 1
    $\begingroup$ @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! $\endgroup$ – SixWingedSeraph Aug 14 '10 at 16:13

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 $.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.