If x is a real number in [0, 1] then the value oflimn→∞ limm→∞ 1+cos2m⁡(n!πx) is given by

If $x \in Q$ then $n! π x$ will be an integral multiple of $π$

for large values of $n$ . Therefore, $\cos (n! π x)$ will be either 1 or

-1 and so $\cos^{2 m} (n! π x) = 1$

$lim_{m \to \infty} lim_{n \to \infty} [1 + \cos^{2 m} (n! π x)] = 1 ∣ + 1 = 2$

If $x \to Q, n! π x$ will not be an integral multiple of $π$ and so

$\cos (n! π x)$ will lie between -1 and 1

thus, $lim_{m \to \infty} \cos^{2 m} (n! π x) = 0$

$\Rightarrow lim_{m \to \infty} lim_{n \to \infty} [1 + \cos^{2 m} (n! π x)] = 1 + 0 = 1 .$

Thus, option (a) is correct.

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