How can a function have a vertical tangent and be continuous? I read somewhere that, "a function with a vertical tangent may be continuous but not differentiable."
Is this correct and, if so, what is an example of it?
I can't think how a function with an asymptote can be continuous.
 A: The idea is just that even though the slope is vertical, you can still approach your point when getting close enough ; the fact that the slope is vertical doesn't mean your function goes to infinity. Among well-known examples, $f(x) = \alpha x^{1/n}$ with $n \ge 1$.  
Hope that helps,
A: As described above, $f(x) = x^{1/3}$ is continuous at $0$ and has a vertical tangent at $0$.
To understand this example more fully, it may be helpful to recall the epsilon-delta definition of "continuous at $0$".
Loosely paraphrased, "$f$ is continuous at $0$" means that for every $\epsilon>0$, we can find $\delta>0$ such that inputs within $\delta$ of $0$ will "force" the outputs to be within $\epsilon$ of $0$.
In the example of $f(x)=x^{1/3}$ at $x=0$, it is true that for every $\epsilon$ there exists a $\delta$. For example, if $\epsilon = 0.01$, we can choose $\delta = 0.000001$. Similar statements can be made for any $\epsilon>0$.
All that matters is that we can find a $\delta$ for every $\epsilon$. But $\delta$ is allowed to be much smaller than $\epsilon$. It might happen that $\delta$ is not a linear function of $\epsilon$, but instead approaches $0$ much "faster" than $\epsilon$ does.
A: Adding a graphic to yoyo's comment:

$\sqrt[3]{x}$ is continuous, but its derivative, $\dfrac{1}{3\sqrt[3]{x^2}}$, tends to $\infty$ near $0$.
