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)?
That fraction has both the numerator & the denominator dependent upon "x"...
I would have used x/x^2, but I wanted to make sure the numerator didn't have a limit in the extended reals either.
