If x is a real number in [0, 1],then the valu | vedclass

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(B) Let $L = \lim_{m 	o \infty} \lim_{n 	o \infty} [1 + \cos^{2m}(n! \pi x)]$. If $x$ is a rational number,then $x = \frac{p}{q}$ for some integers

Vedclass · Accepted Answer

$2$ or $1$ according as $x$ is rational or irrational

Vedclass · Answer

(B) Let $L = \lim_{m 	o \infty} \lim_{n 	o \infty} [1 + \cos^{2m}(n! \pi x)]$. If $x$ is a rational number,then $x = \frac{p}{q}$ for some integers $p, q$. For $n \ge q$,$n! x = n! \frac{p}{q}$ is an even integer (since $n!$ contains $q$ as a factor). Thus,$\cos(n! \pi x) = \cos(k \pi) = \pm 1$ for some even integer $k$,so $\cos^2(n! \pi x) = 1$. Then $\lim_{n 	o \infty} \cos^{2m}(n! \pi x) = 1^m = 1$. So,$L = 1 + 1 = 2$. If $x$ is an irrational number,then $n! x$ is never an integer for any $n$. Thus,$|\cos(n! \pi x)| So,$L = 1 + 0 = 1$. Therefore,the value is $2$ if $x$ is rational and $1$ if $x$ is irrational.

If $x$ is a real number in $[0, 1]$,then the value of $\lim_{m \to \infty} \lim_{n \to \infty} [1 + \cos^{2m}(n! \pi x)]$ is given by

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