Reasoning Behind Holes in Rational Functions I am having some confusion about holes in rational functions. As I'm aware, a hole is where both the numerator and denominator become zero due to some discontinuity. For example, 
f(x) = (x+1)(x-1)/(x+1)
would have a hole at x = -1. 
What is the point of distinguishing between a hole and Vertical Asymptote? 
What is the logic behind the numerator needing to having a discontinuity at the same point as the denominator? 
 A: Both the numerator and the denominator being zero is a necessary but not sufficient condition for a hole; see for example the function $f(x) = \frac{x+1}{(x+1)^2}$. The difference between a hole and a vertical asymptote is that the function doesn't become infinite at a hole. See it for yourself: Take your $f$, evaluate it at points close to $x=-1$ and compare to what happens with mine.
A: Plot the functions $\frac{x^2-1}{x-1}$ and $\frac{1}{x-1}$. You can easily see the difference between a hole and vertical asymptote. A rational function with a hole means it looks very nearly to be a polynomial except that at one (or more points), it is undefined (recall $\frac{0}{0}$ isn't defined). At a vertical asymptote, the function blows up because the denominator approaches zero at some point but the numerator does not (or if it does, not to the same "degree" as the denominator; see Javier's example).
A: A vertical asymptote occurs at a point where the value of the function approaches $\infty$ or $-\infty$.  At a "hole" it doesn't do that.
Some points where the numerator and denominator are both $0$ are holes, but if you reduce the fraction to lowest terms, in some cases you get something where the numerator is not $0$ and the denominator is, so it's a vertical asymptote rather than a hole.
