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

[flying2000]

I can't apply apply L'Hopital here, since the limit of |cosx| doesn't exist..
Any hints appreciated..

[Hurkyl]

Well, just how badly does the limit of ln |cos x| not exist? How do you think your fraction looks as x grows large?

[Zurtex]

Not mathematics in anyway at all:

But wouldn't it be intuitive that this has no limit as as Cos(x) approaches 0, ln(|Cos(x)|) approaches infinity at a faster then exponential rate while 1/(x^2) approaches 0 at a rather slow rate.

[flying2000]

How can I explain that:
ln|cosx| grows faster than x^2 if x>X for some X?

Not mathematics in anyway at all:

But wouldn't it be intuitive that this has no limit as as Cos(x) approaches 0, ln(|Cos(x)|) approaches infinity at a faster then exponential rate while 1/(x^2) approaches 0 at a rather slow rate.

[DeadWolfe]

I think you can use the squeeze theorem to find the limit, since:

0 < |cos x| < 1

from there, if you apply the log and x^-2 functions, then use l'Hospital, I think you could find the answer.

[dextercioby]

I think you can use the squeeze theorem to find the limit, since:

0 < |cos x| < 1

from there, if you apply the log and x^-2 functions, then use l'Hospital, I think you could find the answer.

It's useless,because

\lim_{x\searrow 0} \ln x =-\infty

\lim_{x\rightarrow 1} \ln x =0

Daniel.

[Hurkyl]

How can I explain that:
ln|cosx| grows faster than x^2 if x>X for some X?

ln|cos x| doesn't grow as x grows... you need to understand how it behaves.

[dextercioby]

Here's a plot to convince yourself.

Daniel.

[Data]

How could there be a limit? What's \lim_{u\rightarrow -\infty} \frac{u}{x^2} for any constant x? This is basically an equivalent problem.

[dextercioby]

What do u mean "equivalent problem"...?:confused:That fraction has both the numerator & the denominator dependent upon "x"...

:bugeye:

Daniel.

[Data]

well, \ln{|\cos{x}|} goes to -\infty at every odd multiple of \frac{\pi}{2}. For x > 0, which is obviously reasonable in this case, the largest x^2 gets on any interval \left[k\frac{\pi}{2}, \ (k+2)\frac{\pi}{2}] for an odd integer k>0 is just (k+2)^2\frac{\pi^2}{4} since x^2 is monotonically increasing there.

Thus on any such interval

\left|\frac{\ln{|\cos{x}|}}{x^2}\right| \geq \left|\frac{\ln|\cos{x}|}{u^2}\right|

for some constant u, and from there it's easy.

[dextercioby]

I couldn't follow your logic (it may be my fault),but that limit doesn't exist,because the numerator doesn't have a limit.The denominator goes to +\infty but that still doesn't help.

Daniel.

[Data]

Showing that the fraction goes to -\infty on any interval of the form I posted above is enough (since then you can make x arbitrarily large, and it will still go to -\infty somewhere past that, and in fact (necessarily) infinitely many times).

[dextercioby]

Nope,it can't be -\infty altogether,because u'd have a -\frac{\infty}{\infty} at certain points (a infinite discrete set,where the "cosine=0") and 0 in the other points....

Daniel.

[Data]

Ooops, somehow the \infty on the bottom of your fraction didn't show up at first.

Anyways, you never get an indeterminate form like that. Look at the intervals I was examining. They are all finite, ie. x and x^2 are bounded on each of them.

[Hurkyl]

I couldn't follow your logic (it may be my fault),but that limit doesn't exist,because the numerator doesn't have a limit.

Not a valid reason. For example, consider (cos x) / x^2

[dextercioby]

Yes,but that doesn't help.You need to compute the limit x\rightarrow \pm \infty ,where it doesn't matter whether x^{2} is bounded on an finite interval.

Daniel.

[dextercioby]

Not a valid reason. For example, consider (cos x) / x^2

I agree for the general case.In this context it's valid,though,because the function in the numerator in not bounded.

Daniel.

[Hurkyl]

Nope. Consider (x sin x) / x^2

[Data]

The definition of a limit at infinity is

\lim_{x \rightarrow \infty} f(x) = L \Longleftrightarrow \exists N \ \forall \epsilon > 0 \ \mbox{s.t.} \ x > N \Longrightarrow |f(x) - L| < \epsilon

Now choose any N. I can always show you a point x^\prime with x^\prime>N such that

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

which, from the definition, obviously means that there can't be a finite limit L as x \rightarrow \infty.

[dextercioby]

That's faulty.In the +\infty ,you could simplify the numerator & the denominator through "x"--------->numerator is finite.

I think u would have meant

\lim_{x\rightarrow +\infty} \frac{P(x)\sin x}{Q(x)}

,where P & Q are arbitrary polynomials (for which P(x) doesn't divide Q(x)) with real coefficients and degree of P(x) is stricly less than degree of Q(x).

Daniel.

[dextercioby]

The definition of a limit at infinity is

\lim_{x \rightarrow \infty} f(x) = L \Longleftrightarrow \exists N \ \forall \epsilon > 0 \ \mbox{s.t.} \ x > N \Longrightarrow |f(x) - L| < \epsilon

Now choose any N. I can always show you a point x^\prime with x^\prime>N such that

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

which, from the definition, obviously means that there can't be a finite limit L as x \rightarrow \infty.

That limit (the OP's) is NOT -infty.It doesn't exist.Period.

Daniel.

[Data]

I didn't say it was. Saying that a limit is -\infty (which it isn't in this case) is synonymous with saying the limit doesn't exist as long as you're talking about the real numbers anyhow.

[Hurkyl]

That's faulty.In the +&infin;,you could simplify the numerator & the denominator through "x"--------->numerator is finite.

I agree that it can be simplified. Yet, it still stands as a counterexample -- the numerator is neither bounded, nor does its limit exist as x goes to +&infin;. :tongue2: 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.

[dextercioby]

Yes,Hurkyl,i realized it was a faulty argument.:redface:

There's another one

\lim_{x\rightarrow +\infty} \frac{\ln x\cdot \sin x}{x^{2}} =0

Daniel.

[Zurtex]

So anyway I was thinking about this, why not just apply a simple proof by contradiction.

Assume:

\lim_{x \rightarrow \infty} \frac{ \ln | \cos x | }{x^2} = \alpha \quad \text{for} \, \alpha \in \mathbb{R}

Therefore there exists some x > X such that:

\forall x > X \, : \, \left| \frac{ \ln | \cos x | }{x^2} - \alpha \right| < \varepsilon

Let \varepsilon = 1

Therefore:

\forall x > X \, : \, \left| \frac{ \ln | \cos x | }{x^2} - \alpha \right| < 1

Or:

\forall x > X \, : \, \left| \frac{ \ln | \cos x | - \alpha x^2 }{x^2} \right| < 1

Rewriting further:

\forall x > X \, : (\alpha - 1)x^2 < \ln | \cos x | < (\alpha + 1)x^2

Simply the top limit does not hold true for any large x as x \rightarrow (2k + 1/2)\pi \quad k \in \mathbb{N}. Food is up so I need to go but it shouldn't be too difficult from there, I was thinking to start off by multiplying the whole thing by 2 to get rid of that nasty modulus.

[Zurtex]

Erm Data you seemed to have deleted your reply that that was basically what you were saying, I know that was what you were trying to get at but it didn't seem very rigorous or clearly explained.

[Data]

Yeah it is basically the same I guess, but I decided it was different enough that I shouldn't risk confusing anyone :smile:

My argument is rigorous enough, but I definitely didn't write it out in any complete form.

Yours is a perfectly good (and probably more clear) approach~

[Zurtex]

So anyway I was thinking about this, why not just apply a simple proof by contradiction.

Assume:

\lim_{x \rightarrow \infty} \frac{ \ln | \cos x | }{x^2} = \alpha \quad \text{for} \, \alpha \in \mathbb{R}

Therefore there exists some x > X such that:

\forall x > X \, : \, \left| \frac{ \ln | \cos x | }{x^2} - \alpha \right| < \varepsilon

O.K so take my proof from here and continue:

Let (as if alpha exists it's obviously negative):

\varepsilon = \frac{-1}{(\alpha - 1)}

Then rewriting (and taking a few steps I did from my earlier post):

\forall x > X \, : -x^2 < \ln | \cos x | < \frac{(\alpha + 1)x^2}{\alpha -1}

Looking at the middle term, for x \neq (k + 1/2)\pi \quad k \in \mathbb{N}

\ln | \cos x | = \ln (1 + (| \cos x | - 1)) = \sum_{n=1}^{\infty} \left( \frac{(-1)^{n+1}}{n} (| \cos x | - 1)^n \right)

Now as take some large value of x and increase it so: x \rightarrow (k + 1/2)\pi \quad k \in \mathbb{N}:

| \cos x | - 1 \rightarrow -1

And therefore the sum approaches:

\sum_{n=1}^{\infty} \frac{(-1)^{2n+1}}{n} = -\sum_{n=1}^{\infty} \frac{1}{n}

So it stands that \ln | \cos x | \rightarrow -\infty for some x in the interval [x, \, x+\pi]. Therefore there exists no x > X such that \forall x > X \, : \, -x^2 < \ln | \cos x |.

Contradiction!

Haha, that was a pain, I went down totally the wrong path to start off with and it could be more rigorous but it's pretty good as it is. Well that was certainly good practise for my sequence and series exam in 6 weeks :)

Yeah it is basically the same I guess, but I decided it was different enough that I shouldn't risk confusing anyone :smile:

My argument is rigorous enough, but I definitely didn't write it out in any complete form.

Yours is a perfectly good (and probably more clear) approach~
Your arguement was right but there was no proof to back it up and therefore no good in mathematics.

[Data]

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.

[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! :smile:)

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 :biggrin:


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.