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?

Let $x_n = \frac{1}{\pi(\frac{1}{2}+n)}$, $n=0,1,...$. Note that $\sin \frac{1}{x_n} = (-1)^n$. 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).
