# Question about Continuous functions ($y=\tan x$ example)

I have a question about continuous functions. By definition, a function is said to be continuous in an interval $(a,b)$ if it is continous at any point $x\in(a,b)$. The function $y=\tan x$ is therefore not continous on $\mathbb{R}$ since it is not defined for example at the value $\pi/2$. My question is, If by definition the function has a domain which excludes all those values for which it is not defined, does it remain discontinous?

In the case of $\tan x$, it is continuous wherever it is defined. We do not consider continuity at the points (let one such point be $a$) where the function can't be defined, because the condition of continuity says that $\lim_{x \to a} f(x) = f(a)$, but if $f(a)$ doesn't make sense, then this statement is meaningless.
So for example, define a new domain, $S = \mathbb R - \{ n\frac{\pi}2, n \in \mathbb Z\}$. So we have removed all points where $\tan$ is undefined. On $S$, $\tan x$ is well defined, and continuous.
• So, if i have the function $y=tan(3x^2 -5)$ can I say that it is continuous wherever it is defined? – P.D. Jan 11 '17 at 12:09