What's the difference between continuous and piecewise continuous functions? A continuous function is a function where the limit exists everywhere, and the function at those points is defined to be the same as the limit. 
I was looking at the image of a piecewise continuous function on the following page: http://tutorial.math.lamar.edu/Classes/DE/LaplaceDefinition.aspx 
But the image of the function they've presented isn't continuous. As such, I'm confused by what a piecewise continuous function is and the difference between it and a normal continuous function.
I'd appreciate it if someone could explain the difference between a continuous function and a piecewise continuous function. Also, please reference the image of the piecewise continuous function presented on this page http://tutorial.math.lamar.edu/Classes/DE/LaplaceDefinition.aspx .
Thank you.
 A: $\newcommand{\R}{\mathbb{R}}$The notion of piecewise continuity (PWC) is used differently in different contexts. 
a
Often, a function $f:\R\to\R$ is called PWC if it continuous everywhere, but at a finite number of points. 
In the context of Laplace transform and other integral transforms, a function $f$ is said to be PWC if it is continuous on a partition of intervals of its domain and at the boundaries of the intervals the function has well-defined and finite limits.
Definition. [PWC] A function $f:[a,b]\to\R$ is called piecewise continuous (PWC) if there exist $a = x_0 < x_1 < \ldots < x_n = b$ so that


*

*$f$ is continuous on $(x_k, x_{k+1})$ for all $k=0,\ldots, n-1$

*The limits $\lim_{x\to{}x_{k+1}^{-}}f(x)$ and $\lim_{x\to{}x_{k}^{+}}f(x)$ exist and are finite for all $k=0,\ldots, n-1$
According to this definition, function 
$$
 f(x) = \begin{cases}
          0, &\text{ for } x = 0
\\
          \frac{1}{x}, &\text{ for } x{}>{}0
        \end{cases}
$$
defined over $[0, \infty)$, is not PWC according to the second definition although it has only one point of discontinuity.
Additionally, function $f(x)=\tfrac{1}{x}$, $x\in\R\setminus\{0\}$, is not PWC, again because the limits $\lim_{x\to 0^+}f(x)$ and $\lim_{x\to 0^-}f(x)$ are not finite.
A: A piecewise continuous function doesn't have to be continuous at finitely many points in a finite interval, so long as you can split the function into subintervals such that each interval is continuous.
A nice piecewise continuous function is the floor function:

The function itself is not continuous, but each little segment is in itself continuous.
A: A function $f$ is piecewise continuous on an interval $J\subset{\mathbb R}$ if it is continuous apart from a set of isolated points $\xi\in J$ where only the one-sided limits $\lim_{x\to\xi-}f(x)$ and $\lim_{x\to\xi+} f(x)$ exist.
Note that $f(x):=\sin{1\over x}$ $(x\ne0)$ together with $f(0):=0$ does not define a piecewise continuous function on ${\mathbb R}$, even though this $f$ is continuous in the "segments" created by the special point.
A: A piece-wise continuous function is a bounded function that is allowed to only contain jump discontinuities and fixable discontinuities. These functions almost always occur with the inclusion of floor into the regular set of algebraic functions you are used to in calculus. The reason for this is because up until then there are no functions you have encountered containing any form of jump discontinuity of the finite nature. As the other answer here says, each interval is continuous. This is true. However there are levels of piece-wise continuity which simply put mean that the function is differentiatable fully on those continuous intervals n number of times.
