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Please provide some hints as to how to solve questions with double limits such as this:

$$\lim_{m\to\infty}\left[\lim_{n\to\infty}(\cos(m!\cdot \pi\cdot x))^{2n}\right]$$

One of the things I did was convert the original function to: $$e^{n\ln(\cos(m!\cdot\pi\cdot x)^2)}$$ and then change cosine into sine and take $t=1/m$, and try to use $$\lim \frac{\sin(m!\cdot\pi\cdot x)}{m!\cdot\pi\cdot x}$$ but that only messed it up even further. I obviously can't use L'Hopital as not both the numerator and denominator go to zero.

Another thing was to try to use the power series expansion, but that seemed even more complicated as there is still the power of 2n to deal with.

Please help! Thanks.

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Just to clarify, the power 2n is on the cosine function and not the part inside it. So it goes like (cos(m!.pi.x))^(2n). –  user130113 Feb 21 at 2:37
Hint: Look separately at $x$ rational, $x$ irrational. –  André Nicolas Feb 21 at 2:37
But is my approach right though? –  user130113 Feb 21 at 2:41
I do not see a direct way to use your approach. –  André Nicolas Feb 21 at 2:43
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marked as duplicate by Martin Sleziak, Sami Ben Romdhane, mau, heropup, Yiorgos S. Smyrlis Feb 21 at 9:50

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1 Answer

up vote 5 down vote accepted

Hint: Look separately at $x$ irrational, $x$ rational.

If $x$ is irrational, then $\cos(m!\pi x)$ has absolute value less than $1$.

If $x$ is rational, say $\frac{a}{b}$ where $a$ and $b$ are integers, what is the value of $\cos(m!\pi x)$ for large enough $m$?

Detail: Suppose that $x$ is irrational. Fix $m$. Then $m!\pi x$ is not an integer multiple of $\pi$. It follows that $|\cos(m!\pi x)|\lt 1$. Let $c_m=\cos(m\pi x)$. Since $c_m$ has absolute value less than $1$, we have $\lim_{n\to\infty} c_m^{2n}=0$. Thus our double limit is $\lim_{m\to\infty} 0$, which is $0$.

Now suppose that $x$ is rational. Then we can assume that $x=\frac{a}{b}$, where $a$ and $b$ are integers. Without loss of generality we may suppose that $b$ is positive.

Let $m\ge b$. Then $m!x=(b-1)!a$. Thus $m!\pi x$ is an integer multiple of $\pi$. It follows that $\cos(m!\pi x)=\pm 1$, and therefore $(\cos(m!\pi x))^{2n}=1$. Thus for any $m\ge b$, we have $$\lim_{n\to\infty} (\cos(m!\pi x))^{2n}=1.$$ We conclude that $$\lim_{m\to\infty}\left[ \lim_{n\to\infty} (\cos(m!\pi x))^{2n} \right]=1.$$

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Would it still not be anything between 1 and -1? –  user130113 Feb 21 at 4:31
What is "it"? For $x$ rational, we can without loss of generality assume $b\ge 1$. If $m\ge b+2$, the cosine is $1$. –  André Nicolas Feb 21 at 4:35
This is what I have come to conclude: If m=1, x=1, cos(m!πx)^2n= 1 and for any other x=/ an integer, cos(m!πx)^2n=0. Since m=integer always, it comes down to the value of x. Is that correct at all? –  user130113 Feb 21 at 4:47
We are ultimately not interested in $m=1$, but in large $m$. (For $x$ not rational, the size of $m$ does not matter.) You do not mean that $(\cos(m!\pi x))^{2n}=0$, though the limit is $0$ if $x$ is not rational. At a certain point, if you wish, I can flesh out the hints. –  André Nicolas Feb 21 at 4:51
By the way, important, do check the parentheses, do you mean $\cos((m!\pi x)^{2n})$ or $(\cos(m!\pi x))^{2n}$? –  André Nicolas Feb 21 at 4:53
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