Crazy Limits!!! 2

First of all we will solve $\alpha$

Now

Let x = $\frac{p}{q}$

Substituting in $\alpha$

$\displaystyle \alpha = \lim_{m \to \infty} \lim_{n \to \infty} [cos(n!\pi \frac{p}{q})]^{2m}$

Now we observer that

q will cancel out 100% (with one term of n!)

Then we will get

Cos as multiple of $\pi$

And we know that

$\displaystyle cos(n\pi) = \pm1$

But the power is 2m, taking 2 inside we will only get

$cos(whatever) = 1$

And $1^{\infty} = 1$

We finally get

$\displaystyle \boxed{\alpha = 1}$

Now solving for $\beta$

As x is irrational let is be x

$\displaystyle \beta = \lim_{m \to \infty} \lim_{n \to \infty} [cos(n!\pi x)]^{2m}$

Now

Cos(whatever) will oscillate b/w $\pm 1$

As the power is 2m, taking 2 inside

Cos(whatever) will oscillate b/w

0 and 1

Let angle of cos be 'k'

$\displaystyle \lim_{m \to \infty} [cos(k)]^{m} = 0$

Hence we get

$\alpha = 1, \beta = 0$

Finding the area will require calculator which you can do (comes around 63.94)

Final answer comes 67