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For real $x$, let $$f(x)=\lim_{n\to\infty}(\cos(x))^{2n}$$ How to find the continuous points of $f(x)$?

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When $x=k\pi$, $k$ is integer, $cos(x)=\pm1$ $cos^{2}(x)=1$

When $n\to\infty$, $cos^{2n}(x)=0$ except for $x=k\pi$, $k$ is integer.

Hence $f(x)$ is continuous on $(k\pi,(k+1)\pi)$ $k$ is integer.

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Hint: We have $f(x)=0$ except when $x$ is of the form $k\pi$ for some integer $k$.