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.

Hence, the function definition, comes before the property of continuity at that point. The tangent function is hence continuous wherever it is defined. And if the domain can exclude all the points of discontinuity, then we can absolutely say: "the function is continuous everywhere on it's domain".

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.

  • $\begingroup$ So, if i have the function $y=tan(3x^2 -5)$ can I say that it is continuous wherever it is defined? $\endgroup$ – P.D. Jan 11 '17 at 12:09
  • $\begingroup$ Yes,you can, without thinking twice. $\endgroup$ – астон вілла олоф мэллбэрг Jan 11 '17 at 12:11

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