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. 

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 $\displaystyle\sum_{n=1}^\infty\frac{1}{n}$.
