# 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.)

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Well, it's going to be hard for anyone to draw the function in a neighborhood of zero. Certainly the function equals zero there, but can you see how the $\,x\sin1/x\,$ part looks close to zero? A mess to accurately draw it (in fact, impossible), though some approximation can be made –  DonAntonio Jan 27 '13 at 15:17
But still, the function is only almost an essential discontinuity at 0 -- not an actual essential discontinuity -- so shouldn't it still be possible to draw without lifting the pen? –  Ryan Jan 27 '13 at 15:27
The curve from one side of zero to the other has infinite length, I think (based on estimating the lengths of the zigzags as curve hits $1/x$ then $-1/x$ in turn). Maybe one could say that if drawn, either the pen speed becomes infinite, or it takes infinitely long to draw. –  coffeemath Jan 27 '13 at 15:31
@coffeemath Ah, now Spivak's example is more acceptable--thank you. (I will accept your answer if you post it below.) –  Ryan Jan 27 '13 at 15:34
The graph oscillates infinitely many times in any neighborhood of $0$ and the oscillations become arbitrarily small in amplitude, so in order to draw it precisely you need a pen leaving a trace of width $0$. But ink traces of width $0$ are invisible, thus making the graph impossible to draw. QED. :) –  Andrea Mori Jan 27 '13 at 15:35

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.