A function is said to be continuous. Can it still have vertical asymptotes? This is a general question. A function is said to be continuous. Can it still have vertical asymptotes? Looking at the definition of continuity, I would say no. Because near a vertical asymptote x-delta might have an y of close to minus infinity, while x+delta might have a value of near +infinity, for example.
 A: It doesn't make sense to ask whether a function is continuous at a point where it's not defined. So if it has a vertical asymptote, then that's not a point of discontinuity, but rather a point that is not part of the domain.
To make it clear what I mean by "doesn't make sense", the definition of continuity at a point $x$ involves the expression $|f(x) - f(y)|$. If $x$ is such that $f(x)$ doesn't make sense, then neither does that expression, so asking about continuity is meaningless.
A: An important notion not mentioned in any answer so far is compactness. There is a formal definition of this (https://en.wikipedia.org/wiki/Compact_space) which can be applied to any abstract domain that your function might be defined on, but if your domain is a subset of the number line then it's enough to check whether it's bounded (doesn't "stretch off" to infinity) and closed (includes all its "end points"). Other answers have given various examples of functions which were continuous on their domain and still included a vertical asymptote, but that asymptote was always an "end point" of the domain that was not included in the domain itself; so the domain was not closed and hence not compact.
In general, the image of a compact region under any continuous function will always be compact; so, if a function has values in the numberline (or any other metric space) then the set of values of the function on any compact region of its domain must be compact, and hence bounded, and there can be no vertical asymptote.
A: The function $f\colon \Bbb R\setminus\{0\}\to\Bbb R$ given by $f(x)=\frac 1x$ is continuous on all of its domain. Yet there is a vertical asymptote.
A: Yes. $1/x$ is continous on (0,1) and has a vertical asymptote at 0.
A: "Because near a vertical asymptote x-delta might have an y of close to minus infinity, while x+delta might have a value of near +infinity, for example."
Then you just have to choose a smaller delta.
Take $f(x) = 1/x$ for example.  It is continuous on all $x \ne 0$. And as $f(0)$ is undefined, it is continuous on all points in its domain.
Pick a point $x_1 > 0$.  Then the $\delta$ you choose must be $\delta < |x_1 - 0|=x_1$.  But that is always possible.  If $\delta < x_1$ then $0 < x_1 - \delta < x_1 < x_1 + \delta$ and if $|x - y| < \delta$ then $0< \frac 1{x+ \delta} < \frac 1y < \frac 1{x-\delta}$ does not have the problem of containing a range of unbounded values.
The same argument holds for $x_2 < 0$.
=====
Practical example:  Let $x_1 = 1/\text{googol} = 10^{-100}$.  Is $f(x) = 1/x$ continuous at $x= x_1$?  $x_1$ is pretty damned close the the assymptote, isn't it.
Let $\epsilon > 0$.  We want to find a $\delta > 0$ so the for all $y$ such that $|y- 10^{-100}| < \delta$ then $|f(y) - f(10^{-100})| = |1/y - 10^{100}| < \epsilon$.  
To find such I need $-\epsilon < 1/y - 10^{100} < \epsilon$ or $10^{100} - \epsilon < 1/y < 10^{100} + \epsilon$
or $\frac 1{10^{100}+\epsilon} < y < \frac 1{10^{100} - \epsilon}$.
$\frac 1{10^{100} +\epsilon} -x_1 < y-x_1 > \frac 1{10^{100} - \epsilon}-x_1$
So for $\delta = \min (|10^{100} - \frac 1{10^{100} +\epsilon},|\frac 1{10^{100} -\epsilon} - 10^{100}|)$, as long as $|x_1 - y| < \delta$ then $|1/x_1 - 1/y| < \epsilon$.
Note: $0 < \delta < 1/10^{100}$
So $f$ is continuous at $x = 1/\text{googol}$.  But the delta we had to find was ever smaller than $1/\text{googol}$.
A: The standard definition of continuity only considers points in the domain of the function. Note that by common understanding, a point where a function is undefined, like a vertical asymptote, is not included in its domain. Therefore, a function can have a vertical asymptote and still be a continuous function. 
For example, from Stein and Barcellos, Calculus and Analytic Geometry, 5th Edition (sec. 2.8):

Definition Continuous function. Let f be a function whose domain is the x axis or is made up of open intervals. Then f is a
  continuous function if it is continuous at each number a in its domain.
EXAMPLE 1 Use the definition of continuity to decide whether $f(x) = 1/x$ is continuous.
SOLUTION [reasoning about definition]... Thus $1/x$ is continuous at every number in its domain. Hence it is a continuous function.

Even though, of course, $1/x$ has a vertical asymptote at $x = 0$. 
A: How about $f(x) = x^{1/3}$ (real branch)?  The asymptote in $(0,0)$ is vertical and the function is continuous on the reals.
