Drawing by lifting pencil from paper can still beget continuous function. From page 105 of the 1994 edition of Spivak's Calculus:

A continuous function is sometimes described, intuitively, as one
  whose graph can be drawn without lifting your pencil from the paper.
  Consideration of the continuous function    $$
f(x) =
\begin{cases}
x \sin \frac 1x, & \text{if }x\neq 0 \\
x, & \text{if }x=0
\end{cases}
$$   shows that this description is a little too
  optimistic.

What does Spivak mean? $f(x)$ can be drawn without lifting the pen, can't it?
http://www.wolframalpha.com/input/?i=x+sin+(1%2Fx)
(On the other hand, the function $x \mapsto x$ with domain $\mathbb R -\{0\}$ is clearly continuous but can't be drawn without lifting the pen.)
 A: Perhaps one reason why the graph of the function cannot be drawn is that, since its derivative at 0 does not exist, an artist starting out the graph at 0 would not be able to determine the direction in which the first stroke ought to be made.
A: The curve has infinite length between $x=-1/\pi$ and $x=1/\pi$. To see this note that it passes through each point 
$$\left(\frac{2}{(4k+1)\pi},\frac{2}{(4k+1)\pi}\right)$$ just before passing through
$$\left(\frac{2}{(4k+3)\pi},\frac{-2}{(4k+3)\pi}\right).$$
The distance between these two points is at least the absolute value of $\Delta y$, which is
$$(2/\pi)\left[\frac{1}{4k+1}+\frac{1}{4k+3}\right]$$
which as $k \to \infty$ is asymptotic to $\frac{1}{k\pi}.$ [the same asymptotic estimate occurs between the points going with $1/(4k+3)\pi$ and $1/(4k+5)\pi.$]
So by limit comparison with the sum $\sum 1/(k\pi)$, and the fact that the length computed along straight line segments is less than the curve length, we see there is indeed infinite length as claimed.
This means when drawn the pen point would have to move with unbounded speed, or else that it would take an infinitely long time to draw.
