Let $f:[0,1]\rightarrow \mathbb{R}$ by defined:
$$
f(x)=
\begin{cases}
xsin(\frac{\pi}{2}+\frac{\pi}{x}), x\neq 0\\
0, x=0\\
\end{cases}
$$and let define the curve $\alpha(t)=(t,f(t))$ when $t\in [0,1]$ is the curve rectifiable?
How should we approach this kind of question?
Best Answer
The curve is not rectifiable if we can find a "zig-zag" approximation for the curve with arbitrarily large arc length.
Pick $n\in \mathbb{N}$ greater than $3$, and for each $0< i < n$, let $t_i = \frac{1}{i}$, and let $t_n = 0$, $t_0=1$. Then if $0<i<n$ we have $$f(t_i) = t_i\sin\left(\frac{\pi}{2}+\frac{\pi}{x}\right) = t_i \sin\left(\frac{\pi}{2}(2i+1)\right) = (-1)^i t_i$$ Essentially, this path starts at $1$, goes to each local max and min of $f$ ($n-2$ times), and ends at $0$. Further, the length of this path (if it is defined) satisfies $$L \geq \sum_{i=1}^{n-2} \sqrt{(t_i-t_{i+1})^2+(f(t_i)-f(t_{i+1}))^2} \geq \sum_{i=1}^{n-2} \vert f(t_i) - f(t_{i+1})\vert$$ $$ = \sum_{i=1}^{n-2} \vert t_i+t_{i+1} \vert = \sum_{i=1}^{n-2} t_i+t_{i+1} \geq \sum_{i=1}^{n-2} \frac{1}{i}$$ Taking $n$ to infinity shows us that $L$ is greater than any real number, i.e. the curve is not rectifiable.