I never understood the phrase "continuous on its domain."
Isn't everything continuous on its own domain, since the domain are all the $x$ values that we can plug into $f(x)$ and get a defined $y$ value back? i.e. doesn't the domain by definition tell you where the function is continuous? Why would the domain ever include something not continuous / not defined?
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Well not exactly, because usually when you talk about the domain of the function, you are talking about the natural domain of the function, in other words, where the function is defined, but not necessarly continious. For example $f(x) = sign(x)$ is defined for all $x \in R$, but is not continious on $x = 0$, then is not continious on it´s domain. Actually you can take any step defined function such that is defined on a certain domain but is not continious on that set.
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Definition. A function $f$ is continuous on its domain $D$ if $f$ is continuous at every point $c\in D$.
Example 1. The function $f(x)=e^x$ (with domain $\mathbb R$) is continuous on its domain.
Example 2. Here is a function (with domain $\mathbb R$) that is nowhere continuous: the Dirichlet function $D(x)$. This function can be defined as $$ D(x)= \begin{cases} 1 & \text{ if $x$ is rational } \\ 0 & \text{ if $x$ is irrational. } \end{cases} $$ These examples show that the words "continuous on its domain" are not superfluous.
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A function is said to be continuous if it continues at each point. This means that over the domain. Functions that are not continuous do not exist for every x value over the domain. For example if a function is defined near an open interval (the circle that is not shaded on a graph) then the function is discontinuous. However, if a function is defined near a closed circle (the shaded circle of a graph) then that function is continuous. This problem usually occurs when trying to find the Domain and Range of a function or if a problem ask to graph the Domain and Range of a function.
Not necessarily. The domain of a function tells you over what values the function $f(x)$ exists, not where it is continuous. Take the piecewise function:
$$f(x) = \begin{cases} 1 & x<0\\ 2 & x\geq0 \end{cases}$$
This function is defined for all $x\in\mathbb{R}$, but is not continuous at $x=0$. It still has a valid value: $f(0)=2$, but that doesn't make it continuous at that point.
For a function to be continuous at a point, its limit must be the same regardless of what direction of approach. In this case, $\lim\limits_{x\to0^-}{f(x)}=1$ while $\lim\limits_{x\to0^+}{f(x)}=2$, making it discontinuous at that point.