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Question Number 175147 by infinityaction last updated on 20/Aug/22

if x is a real number in [0, 1]

then the value of

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

Answered by floor(10²Eta[1]) last updated on 21/Aug/22

since \cos (x) is limited then

- 1 ⩽ \underset{n \to \infty}{\lim c o s} (n! π x) ⩽ 1

\Rightarrow 0 ⩽ {\underset{n \to \infty}{\lim c o s}}^{2 m} (n! π x) ⩽ 1

{\underset{x \to \infty}{\lim a}}^{x} = 0 where 0 ⩽ a < 1

\Rightarrow \lim_{m \to \infty} [{\underset{n \to \infty}{\lim c o s}}^{2 m} (n! π x)] = 0 when \cos (n! π x) \neq 1

\cos (n! π x) = 1 \Rightarrow n! π x = 0 \Rightarrow x = 0

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

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