Pretty much any time you can divide by zero, you'll get a discontinuity. Some examples of discontinuous trig functions are tangent, cotangent, secant, and cosecant. If you put them in terms of sine and cosine, you can see that you're dividing by sin(x) or cos(x), which can result in divide by zero.
Piecewise functions are combinations of two different functions over different ranges, and can very easily be discontinuous. According to Wikipedia, this is usually a
jump discontinuity.
If you're looking for a vertical asymptote with a piecewise function, you can try something like log(x) for x>0, log(-x) for x<0.
Um... an exponential function with a negative number as a base has infinitely many removable discontinuities. For example, f(x) = (-5)^x has infinitely many discontinuities. Like, (-5)^ (1/3) is the cube root, which has a real answer (along with two imaginary answers), and (-5)^(1/2) is the square root, which has no real answers. Whenever there's no real answer, there's a gap in the graph. So for rational values of x (those that can be written as a division of two integers, i.e. a fraction), the ones with even denominators will have no real roots, while those with odd denominators will have one real root. I don't know what happens when x is irrational.
The gamma function is another one with infinitely many discontinuities (with vertical asymptotes). You can think of it as an extension of the factorial function to real numbers (which includes non-integers). The difference from my previous example is that it the gamma function is piece-wise continuous, so to speak. It is only discontinuous at each negative integer and zero, so you can break it into a piecewise function. With (-5)^x, on the other hand, no matter how small of a range of x values you pick, you'll find infinitely many discontinuities in that range.
Are discontinuous functions always rational functions?
