# Thread: tough limit

1. $\displaystyle \lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \begin{cases} 1&x \in \mathbb{Q} \\0& x \not\in \mathbb{Q}\end{cases}$. How to prove it?

2. Originally Posted by Lisa91
$\displaystyle \lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \left\{\begin{array}{l} 1 x \in \mathbb{Q}\\1 x \not\in \mathbb{Q}\end{array}\right.$. How to prove it?
Hi Lisa91,

I think there is a typo in the right hand side of the equation. What do you mean by $$1x$$? Is it,

$\displaystyle \lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \left\{\begin{array}{l} x \in \mathbb{Q}\\1 \not\in \mathbb{Q}\end{array}\right.$

Kind Regards,
Sudharaka.

3. Originally Posted by Lisa91
$\displaystyle \lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \left\{\begin{array}{l} 1 \in \mathbb{Q}\\ 0 \not\in \mathbb{Q}\end{array}\right.$. How to prove it?
With great probability Lisa intends the Diriclet function defined as in...

$\displaystyle D(x)= \lim_{m \rightarrow \infty} \lim_{n \rightarrow \infty} \cos^ {2 n} (m!\ \pi\ x) = \left\{\begin{array}{l} 1 \in \mathbb{Q}\\ 0 \not\in \mathbb{Q}\end{array}\right.$ (1)

The 'proof' is relatively easy because if $x \in \mathbb{Q}$ then $x=\frac{p}{q}$ with p and q integer coprimes. Now if $m \rightarrow \infty$ for a certain $m>m_{0}$ q divides m! and it will be $\cos (m!\ \pi\ x) = \pm 1$. Anyway I personally have more than one doubt on the 'logical architecture' of the definition of the Diriclet Function...

Kind regards

$\chi$ $\sigma$

$\displaystyle \lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \begin{cases} 1&x \in \mathbb{Q} \\0& x \not\in \mathbb{Q}\end{cases}$

No, this '2k' has to be in the place I wrote.

5. Originally Posted by Lisa91
$\displaystyle \lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \begin{cases} 1&x \in \mathbb{Q} \\0& x \not\in \mathbb{Q}\end{cases}$

No, this '2k' has to be in the place I wrote.
Also in this case if x is rational for any $n>n_{0}$ the term $n!\ x$ is an even integer so that for any k is $\cos \{(n!\ \pi\ x)^{2k}\}=1$. The problem however is when x is irrational because in this case [probably...] $\cos \{(n!\ \pi\ x)^{2k}\}$ has no limits in n and k...

Kind regards

$\chi$ $\sigma$

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