TITLE: Vertical asymptote of $\frac {3x^2 - 18x - 81}{6x^2 - 54}$
QUESTION [5 upvotes]: Vertical asymptote of  $f(x) = \frac {3x^2 - 18x - 81}{6x^2 - 54}$ is 3, but why not -3?
The original function is already expanded, however, factoring it out a little bit, we get:
$ \frac {3(x-9) (x+3)}{6(x-3)(x+3)}$
The way Salman Khan explained it here didn't really answer that question.
To find vertical asymptote of a rational function we see where it's undefined. And it's undefined when denominator is equal to $0$. As you can see from the factored version of the function, x = -3 does make the denominator equal to $0$!

REPLY [4 votes]: A line with equation $x=k$ is an asymptote for a function $f$ when 
$$
\lim_{x\to k^{\pm}} f(x) =\pm\infty,
$$
meaning that at least one of the four combinations occurs.
For rational functions of the form $f(x)=p(x)/q(x)$ where $p$ and $q$ are polynomials, the condition $q(k)=0$ is a necessary condition for $f$ having $x=k$ as asymptote. Indeed, if $q(k)\ne0$, the function $f$ is continuous at $k$, which implies that $\lim_{x\to k}f(x)=f(k)$ exists and is finite.
However, the condition is not sufficient: the simple example of $f(x)=2x/x$ shows this. It's true that the denominator vanishes at $0$, but
$$
\lim_{x\to0}f(x)=2
$$
so none of the four combinations above occurs.
Your case is very similar:
$$
f(x) = \frac {3x^2 - 18x - 81}{6x^2 - 54}=\frac{(x-9)(x+3)}{2(x-3)(x+3)}
$$
so
$$
\lim_{x\to -3}f(x)=\lim_{x\to-3}\frac{x-9}{2(x-3)}=\frac{-12}{-12}=1
$$
and none of the four combinations mentioned at the start occurs.
You can observe that $\lim_{x\to-3}(3x^2 - 18x - 81)=0$, so $\lim_{x\to-3}f(x)$ is in the indeterminate form $0/0$, meaning that you can't draw conclusions about the limit without further work (in this case, factoring out the common factor $x+3$).
It is possible that both numerator and denominator vanish at a point $k$ and still the line $x=k$ be an asymptote for the function. Consider
$$
f(x)=\frac{x^2-3x+2}{x^2-2x+1}=\frac{(x-1)(x-2)}{(x-1)^2}
$$
Then both numerator and denominator vanish at $1$, but
$$
\lim_{x\to1^-}f(x)=\lim_{x\to1^-}\frac{x-2}{x-1}=\infty,
\qquad
\lim_{x\to1^+}f(x)=\lim_{x\to1^+}\frac{x-2}{x-1}=-\infty,
$$
so $x=1$ is an asymptote for $f$.