Does x->infiity (ln |cosx|)/x^2 exist?

[Zurtex]

Well, it's just a different set of assumptions as to what the people I'm talking to already know. For example, you assumed they know the Taylor expansion of \ln(1+x).

The concept of "proof" is rather subjective in general usage anyways.

That said, as I indicated before, I never actually wrote out my whole argument (assuming that anyone interested could fill in the gaps, since they are rather easy) in any sort of linked form, so you're basically right (although if you combined all the arguments I've made with suitable reordering, the completion of the proof is trivial. It was pretty obvious to start with anyways, so whatever). The original poster asked for hints, so that's all I was trying to give.

They know L'Hopital but nothing about Taylor expansions? It's not a different set of assumptions because I haven't assumed anything in my proof other than well defined or known mathematical results.

The 'concept of "proof"' is not subjective at all in mathematics and when I last check that's what this post is about. You made massive assumptions which you could easily show with other examples how they don't generally apply and without rigourus proof of this example your so called 'proofs' mean nothing.

 

[Data]

Here's the proof you wanted (sorry I didn't post earlier, was busy. You seem to have removed your request, but I already had it almost all typed! )

We claim that

\lim_{x \rightarrow \infty} \frac{\ln|\cos{x}|}{x^2}

does not exist (ie. there is no real number equal to the limit) is a theorem.

Proof of theorem:

Recall the definition of a limit at infinity,

\lim_{x \rightarrow \infty} f(x) = L \Longleftrightarrow \exists N\in \mathbb{R} \ \forall \epsilon \in \mathbb{R}^+ \ \mbox{s.t.} \ x > N \Longrightarrow |f(x) - L| < \epsilon

Here let

f(x) = \frac{\ln|\cos{x}|}{x^2}

so, to be explicit, what we need is

\neg \left(\exists L \in \mathbb{R} \ \mbox{s.t.} \ \left(\exists N \in \mathbb{R} \ \forall \epsilon \in \mathbb{R}^+ \ \mbox{s.t.} \ x > N \Longrightarrow |f(x) - L| < \epsilon\right)\right)

Choose any particular \epsilon \in \mathbb{R}^+. Let N be any real number. Take M = |\lceil N \rceil| \pi + \frac{\pi}{2}. Then since \lceil N \rceil \in \mathbb{Z} we have M = \left(k+\frac{1}{2}\right)\pi for some k \in \mathbb{N}, and M > N.

It is well known that \cos{x} is continuous and that \lim_{x\rightarrow (q+1/2)\pi} \cos{x} =0 for q \in \mathbb{Z}, and thus \lim_{x \rightarrow M} |\cos{x}| = 0, or in other words

\exists \delta \forall \epsilon^\prime \in \mathbb{R}^+ \ \mbox{s.t.} \ |x - M| < \delta \Longrightarrow |\cos{x}| < \epsilon^\prime

It is also well known that \lim_{x\rightarrow 0^+} \ln{x} = -\infty, or in other words

\exists \delta^\prime \in \mathbb{R}^+ \forall \epsilon^{\prime\prime} \in \mathbb{R}^- \ \mbox{s.t.} \ 0< x < \delta^\prime \Longrightarrow \ln{x}<\epsilon^{\prime\prime}

Note that it is also well known that x^2 is bounded on any finite real interval thus is so on the interval I = [M-\pi, M], and that x^2\geq 0 \ \forall x\in \mathbb{R}. Also, that 0 \leq |\cos{x}| \leq 1 \ \forall x \in \mathbb{R} and that \ln{x} \leq 0 for every x \ \mbox{s.t.} \ 0 < x \leq 1, and so it is clear that

f(x) \leq 0

for every x in its domain.

From the above results we get

0 \geq \frac{\ln|\cos{x}|}{u^2} = g(x) \geq f(x)

for some u \in I, for every x \in I for which f(x) is defined.

Taking \epsilon^{\prime}=\delta^\prime gives

\exists \delta \in \mathbb{R}^+ \ \forall \epsilon^{\prime\prime} \in \mathbb{R}^+ \ \mbox{s.t.} \ |x - M| < \delta \Longrightarrow \ln|\cos{x}| < \epsilon^{\prime\prime}

or in other words

\lim_{x \rightarrow M} g(x)u^2 = -\infty

and by the properties of limits, \lim_{x \rightarrow M} g(x) = -\infty as well. Since g(x) \geq f(x), replacement of f(x) in the definition of this limit for g(x) immediately yields

\lim_{x \rightarrow M} f(x) = -\infty

then for any L \in \mathbb{R} we note that again by the properties of limits

\lim_{x \rightarrow M} (f(x) - L) = -\infty

or

\exists \delta^{\prime\prime} \in \mathbb{R}^+ \ \forall \gamma \in \mathbb{R}^- \ \mbox{s.t.} \ |x - M| < \delta^{\prime\prime} \Longrightarrow f(x)-L < \gamma

and choosing |\gamma|>|\epsilon| and \delta^{\prime\prime} < |M - N| (which is justified, since if some \delta^{\prime\prime} > |M-N| works, then so does every \delta^{\prime\prime} < |M - N|, as you can confirm on your own if you like) immediately implies that N and L do not satisfy

x > N \Longrightarrow |f(x) - L| < \epsilon.

Since we put no restrictions on either N or L except that they be real, this proves the theorem. QED.

see, trivial


Anyway, I learned l'Hopital's rule long before I learned about Taylor expansions (unproven, but I was still allowed to use it - crazy math professors).

In the context of this forum, the concept of proof is indeed subjective. You do not know what other posters here know and do not know, so an argument that constitutes a proof to you may not to them (hence the "general usage" qualification - in formal mathematics, things are defined before they are used, and proof is no longer nearly so much a subjective notion). See the play "Proofs and Refutations" for an entertaining demonstration.

I also never called anything that I posted previously in this thread a proof.

I am curious though, what other examples show that the reasoning in my other posts doesn't generally apply (I don't really know what this means anyway, since the reasoning was pretty problem-specific)?

 

[Zurtex]

I was unconvinced you knew a rigorous proof as the general principle of a function not having the limit as it approaches infinity because the numerator is unbounded periodically is not true at all. That's what you seemed to be your earlier argument.