For real $x$, let $$f(x)=\lim\limits_{n\to\infty}(\cos(x))^{2n}$$ How to find the continuous points of $f(x)$?
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
Hint: We have $f(x)=0$ except when $x$ is of the form $k\pi$ for some integer $k$. |
|||
|
|
|
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. |
|||
|
|