TITLE: Give an indeterminate limit of a function that is always indeterminate with iterated attempts at l'Hopital's Rule.
QUESTION [5 upvotes]: I am freshening up on differential calculus, and I came up with a thought experiment in the context of using l'Hopital's rule.
Is there a limit of a function that will always give an indeterminate form regardless of the number of times the numerator and denominator are differentiated?
Answerers can choose any value to be approached, and any function. Single-variable and multi-variable are fine. I also don't mind which indeterminate forms are involved.

REPLY [6 votes]: Here is a fairly trivial limit:
$$\lim_{x\to\infty}\frac{e^x}{e^x}=1$$
but it is quite obvious L'Hospital's won't go anywhere.
In general, L'Hospital's rule won't work if the quotient of derivatives is cyclic, basically repeating itself.

REPLY [3 votes]: $$\lim_{x\to\infty}\tanh x = \lim_{x\to\infty} \frac{\sinh x}{\cosh x}=\lim_{x\to\infty} \frac{\exp(x)-\exp(-x)}{\exp(x)+\exp(-x)} =\lim_{x\to\infty} \frac{\exp 2x-1}{\exp 2x+1}$$
LHR ends up being fairly unhelpful for all 3 of these, falling into a loop for the 2nd and 3rd expressions, and not getting any simpler for the 4th.
Additional note: Taking the second expression and applying the substitution $u=\sinh x$ gives the limit described in a comment by David Mitra, namely $$\lim_{u\to\infty}\frac{u}{\sqrt{u^2+1}}$$
which also falls into a loop of period 2.