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

$$\mathrm{if}\mathrm{x}\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number}\mathrm{in}[0,1]$$$$\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}$$$$\underset{m\to \mathrm{\infty }}{\lim }\underset{n\to \mathrm{\infty }}{\lim }[1+{\cos }^{2m}(n!\pi x)]$$

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

$$\mathrm{since}\cos \left(\mathrm{x}\right)\mathrm{is}\mathrm{limited}\mathrm{then}$$$$-1⩽\underset{\mathrm{n}\to \mathrm{\infty }}{\lim cos}(\mathrm{n}!\pi \mathrm{x})⩽1$$$$\Rightarrow 0⩽{\underset{\mathrm{n}\to \mathrm{\infty }}{\lim cos}}^{2\mathrm{m}}(\mathrm{n}!\pi \mathrm{x})⩽1$$$${\underset{x\to \mathrm{\infty }}{\lim a}}^{\mathrm{x}}=0\mathrm{where}0⩽\mathrm{a}<1$$$$\Rightarrow \underset{\mathrm{m}\to \mathrm{\infty }}{\lim }\left[{\underset{\mathrm{n}\to \mathrm{\infty }}{\lim cos}}^{2\mathrm{m}}(\mathrm{n}!\pi \mathrm{x})\right]=0\mathrm{when}\cos (\mathrm{n}!\pi \mathrm{x})\ne 1$$$$\cos (\mathrm{n}!\pi \mathrm{x})=1\Rightarrow \mathrm{n}!\pi \mathrm{x}=0\Rightarrow \mathrm{x}=0$$$$$$$$\mathrm{if}\mathrm{x}\ne 0\Rightarrow \underset{\mathrm{m}\to \mathrm{\infty }}{\lim }\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[1+{\cos }^{2\mathrm{m}}(\mathrm{n}!\pi \mathrm{x})]=1$$$$\mathrm{if}\mathrm{x}=0\Rightarrow \underset{\mathrm{m}\to \mathrm{\infty }}{\lim }\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[1+{\cos }^{2\mathrm{m}}(\mathrm{n}!\pi \mathrm{x})]=2$$

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