Rectifiability of a curve Let $f$ be a function defined on $[0,1]$ by 
$$f(x) = { 0, \text{ if } x = 0} $$
$$f(x) = { x \sin \frac 1 x , \text{ if } 0 < x \leq 1} $$
Prove that the curve $\{(x, f(x)) : x \in [0,1]\}$ is not rectifiable. 
I'm not sure how to approach this. The general idea seems logical, we're proving that the length of the curve is infinite, but the method seems difficult to find. 
 A: The length of a curve is by definition the $\sup$ of the lengths of inscribed chord-polygons. For your curve define
$$x_0:=1,\qquad x_k:={2\over k\pi}\quad(1\leq k< N),\qquad x_N:=0$$
and consider the chord-polygon $\gamma_N$  through the points $\bigl(x_k, f(x_k)\bigr)$ $\ (0\leq k\leq N)$. As $\bigl|\sin{1\over x_k}\bigr|$ is alternatively  $0$ and $1$ one has $|f(x_k)-f(x_{k-1})|\geq x_k={2\over k\pi}$; therefore the individual chords (apart from the first and the last one) have a length $>{2\over k\pi}$. It follows that  $\gamma_N$ has a total length $>{2\over\pi}\sum_{k=2}^{N-1}{1\over k}$. Since the sum is unbounded for $N\to\infty$ the considered curve $\gamma$ is not rectifiable.
A: The arclength $L$ of a curve $(x,f(x))$ from $x=a$ to $x=b$ is defined as
$$L=\int_a^b\sqrt{1+(f'(x))^2}dx.$$
Therefore, in this case, $f(x)=\displaystyle x\sin(\frac{1}{x})$, which implies that
$$(f'(x))^2=\Big[\sin(\frac{1}{x})-\frac{1}{x}\cos(\frac{1}{x})\Big]^2\geq-\frac{2}{x}\sin(\frac{1}{x})\cos(\frac{1}{x})+\frac{1}{x^2}\cos^2(\frac{1}{x}).$$
Hence, if $x\in\displaystyle[\frac{1}{2\pi n+\pi/3},\frac{1}{2\pi n}]$ where $n\in\mathbb{N}$, then
$$(f'(x))^2\geq -2(2\pi n+\frac{\pi}{3})+4\pi^2 n^2\cos^2(2\pi n+\pi/3)=\pi^2n^2-4\pi n-\frac{2\pi}{3}.$$
Therefore, the arclength $L$ of  $(x,f(x))$ from $x=0$ to $x=1$ can be estimated as follows:
$$L=\int_0^1\sqrt{1+(f'(x))^2}dx\geq \sum_{n=1}^\infty\int^{\frac{1}{2\pi n}}_{\frac{1}{2\pi n+\pi/3}}\sqrt{1+(f'(x))^2}dx$$
$$\geq \sum_{n=1}^\infty\int^{\frac{1}{2\pi n}}_{\frac{1}{2\pi n+\pi/3}}\sqrt{1+\pi^2n^2-4\pi n-\frac{2\pi}{3}}dx$$
$$=\sum_{n=1}^\infty\sqrt{1+\pi^2n^2-4\pi n-\frac{2\pi}{3}}\cdot\frac{\frac{\pi}{3}}{(2\pi n)(2\pi n+\pi/3)}.$$
It's easy to see that the last series in $n$ diverges to infinity by using limit comparison test with the harmonic series $\displaystyle\sum_{n=1}^\infty\frac{1}{n}$.
