Why does 0/0 mean there it a hole and not an asymptote? https://www.khanacademy.org/math/algebra2/rational-expressions/rational-function-graphing/v/finding-asymptotes-example
At around 8 minutes he says that we want a number which makes only the denominator zero, not both the numerator and denominator zero. Later on he explains that the number on which it was 0/0 is a hole and not an asymptote.
What is the reason for this? Why are numbers on which the function is 0/0 holes and on which only the denominator is zero are asymptotes?
 A: This isn't necessarily true. For example, $f(x) = x/x^2$ has an asymptote at $x=0$ but is of the form $0/0$. What he probably means is that if you have something like
$$g(x)=\frac{(x-1)(x+1)}{x-1}$$
then the $(x-1)$ cancels from numerator and denominator, in which case the function looks like $g(x) = x+1$ everywhere except $x=1$ where it has a hole (because it isn't defined).
A: You can have a vertical asymptote where both the numerator and denominator are zero.
You don't always have an asymptote just because you have a $0/0$ expression.
Now if the denominator is zero (at the point in question) and the numerator is not zero, then you will have a vertical asymptote. This is exactly what it happening in the video where
$$
\lim_{x\to 3} f(x) = \lim_{x\to 0} \frac{9(x-9)}{x-3}
$$
This limit is $\pm\infty$ (depending on the side and so $x=3$ is an vertical asymptote.
Note also that in the example in the video, you have
$$
\lim_{x\to -3} f(x) = \lim_{x\to -3} \frac{3(x-9)}{x-3} = \frac{3(-12)}{-6}.
$$
Since this limit is not $\infty$ (or $-\infty$) the function does not have a vertical asymptote at $x=-3$.
Finding asymptotes is all about finding limits.
