# Thread: Limit as x goes to infinity on hard problem

1. ## Limit 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?

2. But given any number as $4k\pi,k\in\mathbb{Z}$ your function is $1$ and on any $(4k+1)\pi,k\in\mathbb{Z}$, your function is $0$. How could that limit exists?

Maybe, you need $\lim_{x\to 0}\left[\cos\left(\frac{x}{2}\right)\right]^{x^2}=\exp\left\{\lim_{x\to 0} \frac{\log\left(\cos\frac{x}{2}\right)}{\frac{1}{x ^2}}\right\}$... and L'Hôpital does the job.

3. But $x\to\infty$ abu.

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

5. 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

6. Originally Posted by Helpaguyout
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
Are you sure you copied the question correctly because this limit (going by the definition) does not exist (Try for example to evaluate the function in any number of the form $2(-\pi + 2\pi k)$ with $k\in \mathbb{Z}$ ).

7. Ugnh, I feel stupid. It's supposed to be 2/x, sorry guys

8. Ah, okay, makes sense.

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

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

Where the important fact here is that $\frac{\ln(1-x)}x\to-1$ as $x\to0.$

9. $\lim_{x\to\infty} [\cos(\frac{2}{x})]^{x^2}$

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

therefore

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

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

the limit

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