Still struggling with continuity A function is continuous iff it is continuous at all points in its domain. What confuses me about this definition is that for undefined points in a function, those points are removed from the domain.

Does this mean a function $f(x) = x$ with domain $(-\infty, -10) \cup (3, \infty)$ is continuous even though there's a big blank space between the two pieces?

Does this mean $f(x) = \frac{x-3}{x-3}$ is continuous even though it has a removable singularity at $x=3$?

Would a single point be continuous even though it lacks limits on either side?

Does this mean $f(x) = \tan(x)$ and $f(x) = 1/x$ are continuous even though they have asymptotes where the function goes to infinity?

Does this mean the only functions that are discontinuous are ones with jump discontinuities?
Terence Tao's book Analysis II makes the following definition (p. 420):


  Let $(X,d_X)$ be a metric space, and let $(Y,d_y)$ be another metric
  space, and let $f:X \to Y$ be a function. If $x_0 \in X$, we say that
  $f$ is continuous at $x_0$ iff for every $\epsilon &gt; 0$, there exists
  a $\delta &gt; 0$ such that $d_Y(f(x),f(x_0)) &lt; \epsilon$ whenever
  $d_X(x,x_0) &lt; \delta$. We say that $f$ is continuous iff it is
  continuous at every point $x \in X$.


Folland gives a definition similar to Tao's.

So, using the definition in Tao or Folland, it is correct to say that the functions you mentioned are continuous. For example, using Tao's definition, it is correct to state that the function $f(x) = \tan(x)$ is continuous. (But the Wikipedia article quoted below warns us that not everyone uses Tao's terminology, so we should be careful that the meaning of this statement is clear in context.)

The answer to your final question is no. Picture the function $f$ defined by  $f(x) = \sin(1/x)$ if $x \neq 0$ and $f(0) = 0$.



By the way, we should note that the Wikipedia says the following:


  There are several different definitions of continuity of a function.
  Sometimes a function is said to be continuous if it is continuous at
  every point in its domain. In this case, the function $f(x) = \tan(x)$,
  with the domain of all real $x \neq (2n+1)\pi/2$, $n$ any integer, is
  continuous. Sometimes an exception is made for boundaries of the
  domain. For example, the graph of the function $f(x) = \sqrt{x}$, with the
  domain of all non-negative reals, has a left-hand endpoint. In this
  case only the limit from the right is required to equal the value of
  the function. Under this definition $f$ is continuous at the boundary $x= 0$ and so for all non-negative arguments. The most common and restrictive definition is that a function is continuous if it is
  continuous at all real numbers. In this case, the previous two
  examples are not continuous, but every polynomial function is
  continuous, as are the sine, cosine, and exponential functions. Care
  should be exercised in using the word continuous, so that it is clear
  from the context which meaning of the word is intended.
It is because we get a little bit loosey-goosey with the concept of domain.

We had too many questions in Algebra class / pre-calculs class were we were asked what is the domain of $f(x).$  i.e. the domain of $f(x) = \frac 1x$ is $\mathbb R - \{0\}.$  But, it is really the responsibility of whoever it is that defines the function to also define the domain. It could be that there are points in the domain where the function is not defined.  When we get to these continuity questions, we are often being asked to include points in the domain where the function may, in fact, be undefined.

$f(x) = \frac 1{x}$ is continuous on $(-\infty,0)\cup (0,\infty),$ but it is not continuous on other domains such as $[-1,1]$ or $\mathbb R.$