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I'm looking for a limit of the form $\lim_{x \to ?}\frac{f(x)}{g(x)}$ such that any arbitrary number of iterations of L'Hospital's rule results in an indeterminate form and the limit that could (most easily?) be calculated by taking an infinite number of iterations -- ie calculating $$\lim_{x \to ?} \left(\lim_{n \to \infty} \frac{f^n(x)}{g^n(x)}\right)$$

I've seen questions like When does L' Hopital's rule fail? -- but the examples from this question either don't have an 'infinite derivative' (because it just alternates) or else just obviously simplify (I'm not sure how to formally rule this case out, maybe by requiring the function $\frac{f^n(x)}{g^n(x)}$ to not 'have a hole' at the limit evaluation point for all $n$).

Is this even possible?

The direction of my own work on this problem has been looking for a trig function that alternates between two values when the limit is taken at infinity, and each successive derivative decreases the distance between the alternating values. After an infinite number of derivative operations, the distance between the alternating values would essentially be 0 and a limit at infinity would exist. I believe a ratio of two such functions could satisfy the requirements of my question.

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    $\begingroup$ $\lim\limits_{x\rightarrow\infty }{x\over\sqrt{x^2+1}}$ or $\lim\limits_{x\rightarrow\infty }{e^x\over e^x}$. $\endgroup$
    – David Mitra
    Commented May 29, 2015 at 21:12
  • $\begingroup$ We can usually sneak around difficulties. In he example $\frac{x}{\sqrt{x^2+1}}$, we can square before hospilatizing. $\endgroup$
    – André Nicolas
    Commented May 29, 2015 at 21:16
  • $\begingroup$ or use power series, or... just do it from scratch, $\endgroup$
    – Matematleta
    Commented May 29, 2015 at 21:28

2 Answers 2

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Take $\lim _{x\rightarrow \infty}\frac{x}{(1+x^{2})^{1/2}}$

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  • $\begingroup$ It is impossible to take the 'infinite derivative' of each iteration's numerator/denominator as they just alternate. $\endgroup$
    – Jonny
    Commented May 29, 2015 at 22:49
  • $\begingroup$ right so you have to use another method. i.e L'Hospital won't work. $\endgroup$
    – Matematleta
    Commented May 29, 2015 at 22:55
  • $\begingroup$ Right, but what I want is to be able to get a result by performing an infinite number of L'Hospital iterations. $\endgroup$
    – Jonny
    Commented May 29, 2015 at 22:56
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    $\begingroup$ You are asking for a situation in which an infinte number of L'Hospital applications will give you the limit? Unfortunately, you run into trouble right awaywhen you try to define just what an "infinite" number of operations is. $\endgroup$
    – Matematleta
    Commented May 29, 2015 at 23:00
  • $\begingroup$ Lets say $\omega$-many iterations of the derivative operation. $\endgroup$
    – Jonny
    Commented May 29, 2015 at 23:08
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You can consider the following example: why $e^x$ beats $x^n$ for $\forall n \in \mathbb{N}$?

Using De Hopital you have: $$\lim_{x \rightarrow +\infty} \frac{e^x}{x^n} \stackrel{\text{De Hopital}}{=} \lim_{x \rightarrow +\infty} \frac{e^x}{nx^{n-1}} \stackrel{\text{De Hopital}}{=} \dots \lim_{x \rightarrow +\infty} \frac{e^x}{n! \, x^0} = +\infty $$ In particular you can take limit $n \rightarrow +\infty$.

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  • $\begingroup$ I like it! But this is an infinite schema of expressions, and for each one the rule succeeds. $\endgroup$
    – Jonny
    Commented May 29, 2015 at 23:11

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