Can a graph of a continuous function intersect its vertical asymptote? I know a graph can intersect its horizontal asymptote (1, 2, 3).
I think it's possible for a graph of a function to intersect its vertical asymptote. Example: Define $f:\mathbb{R}\rightarrow \mathbb{R}$ by $f(x)=1/x$ for $x\neq 0$ and $f(x)=0$ for $x=0$. Then $x=0$ is a vertical asymptote for and intersects $f$. (Correct me if this example is mistaken.)
I think though that it's not possible for a graph of a continuous function to intersect its vertical asymptote. Is this true? How do I prove this?
 A: The answer to this question very much depends on definitions.  In particular, what is the codomain of the functions being considered?  In an introductory calculus class, this question would naturally arise in the context of real-valued functions of a real variable.  In that context, the relevant definitions are

Definition:  A function $f : D \to \mathbb{R}$ is continuous at a point $a$ in its domain $D$ if
$$ \lim_{x\to a} f(x) = f(x). $$
(That is, the limit is defined, and the value of the limit is equal to the value of the function).

and

Definition:  A function $f$ has a vertical asymptote at $a$ if
$$ \lim_{x\to a} f(x) = \pm \infty. $$

These two definitions are, in the context of real-valued functions, contradictory.  Suppose that $f$ is continuous at $a$.  Then
$$ \lim_{x\to a} f(x) = f(a) \in \mathbb{R}. $$
It is then impossible to have $\lim_{x\to a} f(x) = \pm \infty$, as neither $\pm\infty\in\mathbb{R}$.  Thus if a function is continuous at $a$, it cannot have an asymptote at $a$.  Conversely, by a similar argument, if $f$ has an asymptote at $a$, then it cannot be continuous at $a$ (either $a$ is not in the domain of $f$, or you run into the same problem of the limit needing to be both infinite and a real number).
It is worth pointing out that a discontinuous function can "cross" a vertical asymptote.  For example, the function
$$x \mapsto \begin{cases}
\dfrac{1}{x} & \text{if $x \ne 0$, and} \\[1ex]
0 & \text{if $x=0$}
\end{cases} $$
has a vertical asymptote at zero, but crosses that asymptote in the sense that there is a point of the function which lies on the asymptote (specifically, the point $(0,0)$).
However, in the study of functions beyond an elementary calculus course, there are contexts in which a function can have an asymptote at a point of continuity.  For example, consider the function
$$ g : \mathbb{R} \to \overline{\mathbb{R}} : x \mapsto \begin{cases} \dfrac{1}{x^2} & \text{if $x \ne 0$, and} \\[1ex]
+\infty & \text{$x = 0$,} \end{cases} $$
where $\overline{\mathbb{R}}$ is the extended real line, defined by $\mathbb{R} \cup \{+\infty, -\infty\}$.  Here, the codomain is replaced by a set which contains two special elements:  positive and negative infinity, which are defined by the property that if $x \in \mathbb{R}$, then
$$ -\infty < x < +\infty. $$
It takes a little bit of work to show that "continuity" still makes sense for functions with this codomain (the relevant field of study is point-set topology), but it can be done.  In this setting,
$$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{1}{x^2} = +\infty = f(0), $$
hence the function both has a vertical asymptote at zero, and is continuous at zero.
A: No.
By definition, a function $f$ has a vertical asymptote at $a$ if $$\lim_{x\to a} f(x) = \pm \infty$$
Using this definition, it can easily be shown that, if a function has a vertical asymptote at $a$ and the function is also defined at $a$, then the function is not continuous. So, the function's graph cannot intersect the vertical asymptote because in order for it to do so, the graph should contain a point on the asymptote.
