# cos(x)/x amplitude vs squeeze

• Nov 24th 2012, 03:18 PM
friesr
cos(x)/x amplitude vs squeeze
I was presented a question on
$\displaystyle \frac{cos(x)}{x}$ as x approaches infinity.

My approach was to separate the amplitude

http://latex.codecogs.com/png.latex?\frac{1}{x} cos(x)

My explination was since the amplitude was going to zero as x approached infinity, the limit was 0.

I was told this is not correct at all, which i fail to understand how my approach is not correct at all. In fact i can not imagine a single situation with cos(x) where if the amplitude is zero, the over all cos function would do anything but go to zero.

The problem, I was told, was that i did not use the squeeze theorem. i was expected to show

http://latex.codecogs.com/png.latex?... < \frac{1}{x}

I do not understand how my approach was not correct at all vs using the squeeze theorem. Anyone explain where my logic if flawed?
• Nov 24th 2012, 03:30 PM
skeeter
Re: cos(x)/x amplitude vs squeeze
technically, your method uses the product of limits ...

$\displaystyle \lim_{x \to \infty} \frac{\cos{x}}{x} = \lim_{x \to \infty} \frac{1}{x} \cdot \lim_{x \to \infty} \cos{x}$

since $\displaystyle \lim_{x \to \infty} \cos{x}$ does not exist, the product of limits does not show that the overall limit is zero ... you would have to use the fact that $\displaystyle -1 \le \cos{x} \le 1$, hence the need for the squeeze theorem.
• Nov 24th 2012, 04:52 PM
Prove It
Re: cos(x)/x amplitude vs squeeze
Quote:

Originally Posted by friesr
I was presented a question on
$\displaystyle \frac{cos(x)}{x}$ as x approaches infinity.

My approach was to separate the amplitude

http://latex.codecogs.com/png.latex?\frac{1}{x} cos(x)

My explination was since the amplitude was going to zero as x approached infinity, the limit was 0.

I was told this is not correct at all, which i fail to understand how my approach is not correct at all. In fact i can not imagine a single situation with cos(x) where if the amplitude is zero, the over all cos function would do anything but go to zero.

The problem, I was told, was that i did not use the squeeze theorem. i was expected to show

http://latex.codecogs.com/png.latex?... < \frac{1}{x}

I do not understand how my approach was not correct at all vs using the squeeze theorem. Anyone explain where my logic if flawed?

Actually your method IS correct, it's just worded in an incorrect way. You need to say that the product of a function that goes to 0 with a BOUNDED function has a limit of 0.
• Nov 25th 2012, 05:41 AM
friesr
Re: cos(x)/x amplitude vs squeeze

Since

$\displaystyle \lim_{x\to\infty} \frac{1}{x}$ is zero and the $\displaystyle \lim_{x\to\infty} Cos(x)$ does not exist. Then $\displaystyle 0 * DNE = DNE$ is the problem with my approach. This is why I would have needed to state that Cos(x) was bounded so that it would have existed or used squeeze on the whole function?

Thanks for clearing that up,
Rod