I know that there's a theorem which says that if $f$ is a continuous function, then: $$\lim f(x_n) = f(\lim x_n)$$ This is used to solve, for example: $$\lim(\sin(\frac{2n\pi}{1 + 8n})) = \sin(\lim \frac{2n\pi}{1 + 8n}) = \frac{\sqrt{2}}{2}$$ My question is: can I use this theorem to functions that are continuous only in their domain (not on the entire real set) ? For example: $$\lim(\tan(\frac{2n\pi}{1 + 8n})) = \tan(\lim \frac{2n\pi}{1 + 8n}) = 1$$ The theorem apply? Can I do this safely ?
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- 3$\begingroup$ A function can't be continuous outside of its domain. $\endgroup$Git Gud– Git Gud2013-06-02 20:50:29 +00:00Commented Jun 2, 2013 at 20:50
- $\begingroup$ Ok, I understand. What I'm asking is if the domain needs to be the real set. $\endgroup$Otávio Rapôso– Otávio Rapôso2013-06-02 20:52:40 +00:00Commented Jun 2, 2013 at 20:52
- 2$\begingroup$ The only domain that you need $\tan$ to be continuous on is an interval surrounding $\frac{\pi}{4}$. $\endgroup$vadim123– vadim1232013-06-02 20:56:14 +00:00Commented Jun 2, 2013 at 20:56
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1 Answer
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No, check this link.
A function $f$ is continuous at $x_0$ in its domain if for every sequence $(x_n)$ with $x_n$ in the domain of $f$ for every $n$ and $\lim x_n = x_0$, we have $\lim f(x_n)=f(x_0)$. We say that $f$ is continuous if it is continuous at every point in its domain.
The limit of the sequence is required to be in the domain of the fucntion. Because, of course, it doesn't make sense to consider $f(x)$ if $x\not \in \operatorname{dom}(f)$.
