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

[flying2000]

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

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]

still not get it..

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"...?That fraction has both the numerator & the denominator dependent upon "x"...



Daniel.

 

[Data]

well, \ln{|\cos{x}|} goes to -\infty at every odd multiple of \frac{\pi}{2}. For x &gt; 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&gt;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 &gt; 0 \ \mbox{s.t.} \ x &gt; N \Longrightarrow |f(x) - L| &lt; \epsilon

Now choose any N. I can always show you a point x^\prime with x^\prime&gt;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 &gt; 0 \ \mbox{s.t.} \ x &gt; N \Longrightarrow |f(x) - L| &lt; \epsilon

Now choose any N. I can always show you a point x^\prime with x^\prime&gt;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;. 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.

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 &gt; X \, : \, \left| \frac{ \ln | \cos x | }{x^2} - \alpha \right| &lt; \varepsilon

Let \varepsilon = 1

Therefore:

\forall x &gt; X \, : \, \left| \frac{ \ln | \cos x | }{x^2} - \alpha \right| &lt; 1

Or:

\forall x &gt; X \, : \, \left| \frac{ \ln | \cos x | - \alpha x^2 }{x^2} \right| &lt; 1

Rewriting further:

\forall x &gt; X \, : (\alpha - 1)x^2 &lt; \ln | \cos x | &lt; (\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

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 &gt; X \, : \, \left| \frac{ \ln | \cos x | }{x^2} - \alpha \right| &lt; \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 &gt; X \, : -x^2 &lt; \ln | \cos x | &lt; \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 &gt; X such that \forall x &gt; X \, : \, -x^2 &lt; \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

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 argument 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.