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

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

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up vote 0 down vote accepted

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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