I need the limit of x to infinity on the problem (cos(2/x))^(x^2) (the quantity cos of 2/x raised to the x squared). My gut told me it would be one, however the true answer is around .1353. How do you solve it?

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- Jan 3rd 2010, 12:09 PMHelpaguyoutLimit as x goes to infinity on hard problem
I need the limit of x to infinity on the problem (cos(2/x))^(x^2) (the quantity cos of 2/x raised to the x squared). My gut told me it would be one, however the true answer is around .1353. How do you solve it?

- Jan 3rd 2010, 02:32 PMAbu-Khalil
But given any number as $\displaystyle 4k\pi,k\in\mathbb{Z}$ your function is $\displaystyle 1$ and on any $\displaystyle (4k+1)\pi,k\in\mathbb{Z}$, your function is $\displaystyle 0$. How could that limit exists?

Maybe, you need $\displaystyle \lim_{x\to 0}\left[\cos\left(\frac{x}{2}\right)\right]^{x^2}=\exp\left\{\lim_{x\to 0} \frac{\log\left(\cos\frac{x}{2}\right)}{\frac{1}{x ^2}}\right\}$... and L'Hôpital does the job. - Jan 3rd 2010, 02:39 PMKrizalid
But $\displaystyle x\to\infty$ abu.

- Jan 3rd 2010, 02:53 PMJose27
I don't believe this question is correct since clearly $\displaystyle \cos \left( \frac{x}{2} \right)$ has negative values in any interval of the form $\displaystyle (a,\infty )$ where $\displaystyle a>0$ therefore the usual definition of $\displaystyle x^y$ does not apply (at least in $\displaystyle \mathbb{R}$ ). Are you by any chance talking about $\displaystyle \mathbb{C}$ ?

- Jan 3rd 2010, 03:13 PMHelpaguyout
I've checked, and the problem is doable. I believe it is done by constructing a new limit then somehow substituting in, but I don't know how to do that

- Jan 3rd 2010, 03:22 PMJose27
- Jan 3rd 2010, 03:32 PMHelpaguyout
Ugnh, I feel stupid. It's supposed to be 2/x, sorry guys

- Jan 3rd 2010, 04:03 PMKrizalid
Ah, okay, makes sense.

We proceed as follows: $\displaystyle \frac{1-\cos \frac{2}{x}}{\frac{1}{x^{2}}}\to2$ as $\displaystyle x\to\infty.$ (Very easy to prove, just put $\displaystyle t=\frac1x.$) So

$\displaystyle \cos ^{x^{2}}\left( \frac{2}{x} \right)=\exp \left( x^{2}\cdot\frac{\ln \left( 1-\left( 1-\cos \frac{2}{x} \right) \right)}{1-\cos \frac{2}{x}}\cdot \left( 1-\cos \frac{2}{x} \right) \right)\to e^{(-1\cdot 2)}=\frac{1}{e^{2}},$ as $\displaystyle x\to\infty.$

Where the important fact here is that $\displaystyle \frac{\ln(1-x)}x\to-1$ as $\displaystyle x\to0.$ - Jan 3rd 2010, 11:02 PMsimplependulum
$\displaystyle \lim_{x\to\infty} [\cos(\frac{2}{x})]^{x^2}$

we know that if $\displaystyle \theta \to 0$ , then $\displaystyle \cos(2\theta) = 1- 2\sin^2(\theta) \to 1 - 2 \theta^2 $

therefore

$\displaystyle [\cos(\frac{2}{x})]^{x^2} \to \left[ 1 - 2(\frac{1}{x})^2 \right ]^{x^2} $

replace $\displaystyle x^2 $ with $\displaystyle n $ , which also tends to infinity , too .

the limit

$\displaystyle = \lim_{n\to\infty} \left ( 1 - \frac{2}{n} \right )^n = e^{-2} $