So I'm studying infinite limits in my calculus text (seemed close enough to good old arithematic to put in general math, though), and the following rule is mentioned: Given two functions f(x) and g(x) defined for all real numbers, when given the quotient [tex]f(x)/g(x)[/tex] where f(c) is not 0 and g(c) is 0, there is a vertical asymptote at c. But for[tex]f(c)/g(c)[/tex] where f(c) = 0 and g(c) = 0 it is not guarenteed that there is a vertical asymptote at c. Now obviously the crappy off-the-top-of-my-head definition isn't the perfect one given in the book, but I'm sure if you're able to help you know the one I'm talking about. Now here's the thing: Every time I've done an exercise that was in indeterminate form for a value c and found analytically for a value of x that isn't defined, I've found that the numerator and denominator have the same factor that can be cancelled out. Consider the following: [tex]f(x) = (x^2 -1)/(x-1)[/tex], when you 'input' f(1) you get 0/0, indeterminate form. But you can factor the equation to [tex] (x+1)(x-1)/(x-1) [/tex], cancel out the x-1, and see that f(1) is really 2 (or, at least, a function agreeing at every point except x = 1 is really 2). This has been my experience with all exercises involving indeterminate form. So my question(s) becomes: if [tex]f(c)/g(c) = 0/0[/tex], then does this imply the existence of like factors in both the numerator and denominator that can be factored out and cancelled out? If so how is this proven? Even a simple why would be much appreciated.