# Showing $\sin(1/x)$ is not a rectifiable curve

Intuitively it looks like near $0$, $\sin(1/x)$ oscillates wildly so that two points will be very far apart, but how can I properly formulate this?

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Probably it is intuitively obvious, and we can leave it at that! But if you want to be more formal, look at the arclength from $0$ to $1$. Set up the integral and show it does not converge. Or even more properly, look at arclength from $\epsilon>0$ to $1$, and show it blows up as $\epsilon$ approaches $0$. So we calculate $\sqrt{1+(f'(x))^2}$. You will see this is real big near $0$. –  André Nicolas Jun 2 '12 at 15:50
Hint: Remember that the supremum is taken over all partitions of an interval, and notice that as $x$ approaches 0, the function will go from +1 to -1 and back an unbounded number of times, so the supremum must also be unbounded. –  Old John Jun 2 '12 at 15:51

Let $x_n = \frac{1}{\pi(\frac{1}{2}+n)}$, $n=0,1,...$. Note that $\sin x_n = (-1)^n$. Note that $x_n$ monotonically decreases to $0$.
Consider the function on the interval $[x_n,x_0]$, using the partition $t_k = x_{n-k}$. Then the variation is exactly $2n$. Hence the variation is unbounded on the interval $(0,1)$ (or any open interval with $0$ as the left most point).