What are the points of discontinuity of $\tan x$? $f(x) = \tan x$ is defined from $\mathbb R - \{\frac{\pi}{2} (2n+1) \mid n \in \mathbb Z\}$ to $\mathbb R$. For every $x$ in its domain,
$$f(x) = \frac{\sin x}{\cos x}$$ where $\cos x$ is never 0. Thus, (in short) $\tan x$ is defined for all points in its domain.
Now the question remains, is $\tan x$ discontinuous at $x = \pi/2$ (which is outside its domain)?
The question arises because the test for continuity in a textbook mentions that $f(x)$ is continuous at $x = c$ when:


*

*$f(c)$ exists.

*$\lim_{x \to c} f(x)$ exists.

*$f(c) = \lim_{x \to c} f(x)$.


And my teacher says failure of any of the above results in $x = c$ being a point of discontinuity. Yet, according to me, first test above merely tests the point for its domain and should be the criteria for any point of discontinuity too.
 A: This is a real nightmare, for university teachers. Every student of mine comes to my class from high school and is sure that $x \mapsto 1/x$ is discontinuous at $0$. The reason why calculus textbooks are so ambiguous is that their authors do not like to leave something undiscussed. For some reason, the answer to any question should be "yes" or "no"; hence they tend to formally state that functions are discontinuous outside their domain of definition.
In my opinion, this is a very bad approach: it is a matter of fact that discontinuous should not be read as the negative of continuous. The domain of definition makes a difference, and the most useful idea is that of continuous extension.
Almost any mathematician would say that the tangent function is continuous inside its own domain of definition. 
A: I am afraid that the definition of the continuity of a function $f$ in given point $x$ requires that function to be defined in $x$. If $x$ is outside of the domain of $f$, then you can say that neither f is continuous in $x$ nor that $f$ is discontinuous in $x$. Definition just do not cover such a situation.
You may also want to consult the Definition of continuity or Classification of discontinuities.
A: $\tan$ is continuous. As others have mentioned, considering continuity outside domain doesn't make sense.
The continuity can be made more apparent if you consider $\tan$ as a restriction of its natural extension to a function from $\mathbf R$ to the one-point compactification of reals $\mathbf R\cup \{\infty\}$. The extension makes full "circles" in a continuous way, and a restriction of a continuous function is again continuous.
In general, quotients of continuous functions are continuous where defined, because the function $\varphi:(x,y)\mapsto x/y$ is continuous for nonzero $y$, so for continuous $f,g$ we can consider $f/g$ as a composition $\varphi\circ (f,g)$, and composition of continuous functions is continuous.
A: There are no “right” or “wrong” definitions, but there are “standard” and “non-standard” definitions. In my opinion, the definition of continuity from your textbook is non-standard when applied to functions defined on sets with isolated points. For example, the function $f:{\mathbb Z} \to {\mathbb R}$ defined by $f(x) = x$ is a continuous function according to the standard definition but it is discontinuous according to the definition from your textbook.
Here is a standard definition.
Let $D \subset {\mathbb R}$ and $f:D \to {\mathbb R}$. We say that $f$ is continuous at point $c$ if one of the following condition holds:


*

*$c$ is a limit point of $D$ and $\lim_{x\to c} f(x) = c$, or

*$c$ is an isolated point of $D$.


Some textbooks define essential discontinuities even for points in $\bar D$ as follows.
Let $E \subset {\bar D}$. We say that $c\in E$ is an essential discontinuity of $f$ on $E$ if there is no function $\hat f: E \to {\mathbb R}$ such that


*

*${\hat f}(x) = f(x)$ for $x\in D\cap E$,

*$\hat f(x)$ is continuous at point $c$.


According to this definition, $\pi/2$ is an essential discontinuity of $\tan x$ on $\mathbb R$. Of course, it's important what the set $E$ is, in this definition. For example, consider the the function $g(x):{\mathbb R} \setminus {\mathbb Z}\to \mathbb R$ defined by $g(x) = x - \lfloor x \rfloor$. Then 0 is not an essential discontinuity of $g(x)$ on $[0, 1)$, nor on $(-1,0]$. But 0 is an essential discontinuity of $g(x)$ on $(-1,1)$.
A: One way to interpret point 1. is to say that if $x=c$ is not in the domain of $f$, then $f$ is not continuous at $x=c$.
