show that the length of the curve $\alpha$ tends to infinity when $N \to \infty$ The following example shows that for any reasonable definition, the length of a curve $C ^0$ in a range defined closed may be unlimited. Let $ \alpha: [0,1] \to \mathbb {R}^2 $ given by
$$\alpha (t) =\left( t, \sin \frac {\pi} {t}\right) $$ if $ t\neq 0 $, and $\alpha (0) = (0.0) $. Shows geometrically that the arc length of the curve portion corresponding to
$ \frac {1} {n +1} \leq t \leq \frac {1} {n} $
is at least $\frac{2}{n + \frac{1}{2}}$. Use this to show that the curve length in the range $\frac {1} {N} \leq t \leq 1$ is greater than $ 2 \displaystyle \sum_{n = 1}^{N} \frac {1}{n +1} $, thus tending to infinity when $ N \to \infty $.
Do Carmo's exercise? But you miss a $t$, the parameter should be 
$$
\alpha(t) = \left(t, t\sin\left(\frac{\pi}{t}\right)\right),
$$
when $t\neq 0$.
Do Carmo's exercise? But you miss a $t$, the parameter should be 
$$
\alpha(t) = \left(t, t\sin\left(\frac{\pi}{t}\right)\right),
$$
when $t\neq 0$.