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

1. Mar 25, 2005

### 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..

2. Mar 25, 2005

### Hurkyl

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

3. Mar 25, 2005

### 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.

4. Mar 25, 2005

### flying2000

still not get it..

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

5. Mar 25, 2005

### 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.

6. Mar 26, 2005

### dextercioby

It's useless,because

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

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

Daniel.

7. Mar 26, 2005

### Hurkyl

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

8. Mar 26, 2005

### dextercioby

Here's a plot to convince yourself.

Daniel.

#### Attached Files:

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9. Mar 26, 2005

### 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.

10. Mar 26, 2005

### dextercioby

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

Daniel.

11. Mar 26, 2005

### 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.

Last edited: Mar 26, 2005
12. Mar 26, 2005

### 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.

13. Mar 26, 2005

### 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).

14. Mar 26, 2005

### 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.

15. Mar 26, 2005

### 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.

16. Mar 26, 2005

### Hurkyl

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

17. Mar 26, 2005

### 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.

18. Mar 26, 2005

### dextercioby

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

Daniel.

19. Mar 26, 2005

### Hurkyl

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

20. Mar 26, 2005

### 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 $x \rightarrow \infty$.

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