Examples of applying L'Hôpitals rule ( correctly ) leading back to the same state? Are there any examples of applying L'Hôpitals rule ( correctly ) that would yield back the same indeterminate form? I.e. examples where using L'Hôpital either does not reduce the situation or even makes it worse?
By correct usage we mean when the conditions to apply L'Hopitals rule are satisfied, in other words not when L'Hopitals rule is incorrectly applied 
 A: If I understand correctly, you're looking for a case where
$$\frac fg=\frac{f'}{g'}\;,$$
so that replacing the limit of one by the limit of the other yields no improvement. We can rewrite this as
$$\frac{g'}g=\frac{f'}f\;,$$
which can be integrated to
$$\log g=\log f+c\;,$$
$$g=\tilde cf\;,$$
so this condition implies that the two functions are actually multiples of each other.
A: Arjang has asked me to post here what I posted in answer to question 59328, so here goes: 


*

*Let $$f(x)={x\over\sqrt{x^2+1}}$$ and try to find $\lim_{x\to\infty}f(x)$, using l'Hopital. We get $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}{1\over{x\over\sqrt{x^2+1}}}=\lim_{x\to\infty}{\sqrt{x^2+1}\over x}$$ Use l'Hopital again, and you get back to $\lim_{x\to\infty}f(x)$. 

*$$\lim_{x\to\infty}{e^{-x}\over1/x}=\lim_{x\to\infty}{e^{-x}\over1/x^2}=\lim_{x\to\infty}{e^{-x}\over1/2x^3}=\dots$$
A: Here's a "naturally occurring" example:
The Taylor expansion of
$$f(x) = \exp\left( -\frac{1}{x^2}\right)$$
$$f(0) = 0$$
is
$$0 + 0x + 0x^2 + 0x^3 + 0x^4 + ...$$
The simplest way I know to obtain these coefficients is to prove by induction that the $n$th derivative of $f(x)$, for $x \neq 0$, is a rational function times $\exp\left( -\frac{1}{x^2}\right)$ (the rational function will vary with $n$), a form that you'd likely guess if you tried computing the first two or three derivatives. It is not very hard to prove the induction step, namely that the $(n+1)$st derivative is a rational function times $\exp\left( -\frac{1}{x^2}\right)$ when $x \neq 0$. Using this result, we can now prove (again, by induction) that each $n$th derivative evaluated at $x = 0$ exists and is equal to $0$. This time, in proving the inductive step, which involves evaluating the $(n+1)$st derivative at $x = 0$, you're going to need L'Hopital's rule to evaluate the limit as $h \rightarrow 0$ of
$$\frac{f^{(n)}(0 + h) - f^{(n)}(0)}{h},$$
where you will be assuming that $f^{(n)}(0 + h) = f^{(n)}(h) = R(h) \cdot \exp\left( -\frac{1}{h^2}\right)$ for some rational function $R(h)$. However, if you start "L'Hopital differentiating" right away (with respect to $h$), you'll find things will spiral out of control. (Try it by using $R(h) = \frac{P(h)}{Q(h)}$.) To fix this, make the variable change $u = \frac{1}{h}$ and then take the limit as $u \rightarrow \infty$ of the $u$-version, using L'Hopital's rule.
A: Here's one L'Hôpital fixed function: $\ \displaystyle\lim_{x\to\infty}\ \frac{e^x}{e^x\pm e^{-x}}\: \leadsto\ \lim_{x\to\infty}\ \frac{e^x}{e^x\pm e^{-x}} $
A: For an example of order $n$ cyclic behavior, let $f$ and $g$ be real and imaginary parts of $\exp(\omega t)$ for $\omega$ an $n$th root of unity with $\Re \omega < 0$, and apply L'Hopital to the $t \to +\infty$ case of 
$(e^t + f(t)) / (e^t + g(t))$.
With $n=3$ ,  
$f(t) = \exp(-t/2) \cos {\frac{\sqrt{3}}{2}}t$
$g(t) = \exp(-t/2) \sin {\frac{\sqrt{3}}{2}}t$
More interesting would be an example in which the limit is not evident prior to use of L'Hopital's rule.  Here the fixed-point $e^x$ predominates and is added to stabilize the order $n > 1$ oscillatory behavior, but this dominance could have been used to see the limit immediately  without L'Hopital.  Cyclic behavior modulo cancellation, where $D^n (f,g) = (hf,hg)$, does allow for more solutions  but then cancellation also is available to simplify the original limit (e.g., in the case of $\sqrt{x^2+1}/x$ one can divide both by $x$ to change from indeterminate 0/0 to 1/1).
