# Give an indeterminate limit of a function that is always indeterminate with iterated attempts at l'Hopital's Rule.

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.

• $\lim\limits_{x\rightarrow\infty}{x\over\sqrt{x^2+1}}$. – David Mitra Jan 7 '17 at 22:47
• L'Hospital's rule is not the alpha and omega of limits computation. It is a dangerous rule, because beginners often forget to check the conditions of application, and when it works, Taylor's formula at order 1 also works. – Bernard Jan 7 '17 at 22:48
• Here's a tangentially related question to think about (which occurred to one of my students a few years ago): Suppose $f$ and $g$ are differentiable at $a$ and $\lim\limits_{x\to a} f(x)/g(x)$ exists. Does it follow that $\lim\limits_{x\to a}f'(x)/g'(x)$ exists? – Ted Shifrin Jan 7 '17 at 23:34
• @TedShifrin Good question! I've never heard l'Hospital's rule stated as a biconditional, so I have an intuition it is not guaranteed. Make it a SE question and link here in the comments. I'd like to find out. – Galen Jan 8 '17 at 2:18
• @Galen: Oh, I know the answer :) – Ted Shifrin Jan 8 '17 at 2:35

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.

• by "derivative" I think you mean "quotient of derivatives" – zhw. Jan 7 '17 at 23:13
• @zhw. Yes... that is what I meant. – Simply Beautiful Art Jan 7 '17 at 23:43
• Funny for me, I encountered this limit in a practice problem today! – Galen Jan 8 '17 at 2:18
• @Galen Hahaha, a comical moment :D – Simply Beautiful Art Jan 8 '17 at 2:21

$$\lim_{x\to\infty}\tanh x = \lim_{x\to\infty} \frac{\sinh x}{\cosh x}=\lim_{x\to\infty} \frac{\exp(x)-\exp(-x)}{\exp(x)+\exp(-x)} =\lim_{x\to\infty} \frac{\exp 2x-1}{\exp 2x+1}$$

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.

• Hm, I wonder if we can construct a L'Hospital's limit problem where after an infinite amount of applications, it appears to approach the right answer explicitly... :D – Simply Beautiful Art Jan 7 '17 at 22:55
• @SimpleArt not sure what you mean by this. there's a case that says repeated LHR on the 4th expression giving $\frac{2^n\exp(2x)-1}{2^n\exp(2x)+1}$ "hints" that the limit ought to be $1$. This is obviously wrong, though (for example, it could be used to imply the limit at $-\infty$ would also be 1 ...) – πr8 Jan 7 '17 at 22:58
• Hm, I guess it was ill-thought comment. :-P – Simply Beautiful Art Jan 7 '17 at 23:00
• @SimpleArt not necessarily - i just don't entirely understand what you mean by it yet. – πr8 Jan 7 '17 at 23:01
• Its ok, was nothing important. Just a strange side-thought. – Simply Beautiful Art Jan 7 '17 at 23:03