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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.

Martin Sleziak
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user130113
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1 Answers1

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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 $$\lim_{m\to\infty}\left[ \lim_{n\to\infty} (\cos(m!\pi x))^{2n} \right]=1.$$

André Nicolas
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  • Would it still not be anything between 1 and -1? – user130113 Feb 21 '14 at 04:31
  • 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 '14 at 04:35
  • 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 '14 at 04:47
  • 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 '14 at 04:51
  • 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 '14 at 04:53
  • I mean the second one: (cos(m!πx))^2n. I clarified that in the comments box. Could you please edit the question, so that it reflects correctly? And why should the value of m matter anyway? Irrespective of what m is, if x is rational, then the inner part of cosine function will be even, wouldn't it be? And as such the limit would be 1 always. Can you shed some more light on the part when x is irrational? – user130113 Feb 21 '14 at 05:04
  • You are not right about independence from $m$ if $x$ is rational, but for large enough $m$ it is $1$. For $x$ not rational, $m!\pi x$ is not an integer multiple of $\pi$, and therefore the cosine has absolute value $c_m\lt 1$. It follows that $\lim_{n\to\infty} c_m^{2n}=0$. Now take the limit as $m\to\infty$. We are taking a limit of $0$'s, which is $0$. – André Nicolas Feb 21 '14 at 05:10
  • That's what I was thinking about the irrational when I first made the comment that x=/ integer, (cos(m!πx))^2n= 0, but you put it very elegantly in mathematical terms. But I am still pretty confused about the dependence on m. Can you please form an answer so that it becomes more apparent? Thanks. – user130113 Feb 21 '14 at 05:17
  • @user130113: I have extended the hint to an answer. – André Nicolas Feb 21 '14 at 05:38
  • This is great stuff. Thanks a lot. – user130113 Feb 21 '14 at 05:41
  • You are welcome. – André Nicolas Feb 21 '14 at 05:41