Given $\gamma : (-1,1) \rightarrow \mathbb{R}^2$ as a parametric curve with $\gamma(t) = ( \ t, \ t^2sin(\frac{1}{t^2}) \ ) $ for $t \neq 0$ and $\gamma(t) = (0,0)$ for $t=0$. Show that $\gamma(t)$ is differentiable but not rectifiable.
I already showed that $\gamma(t)$ is differentiable. But how to show that $\gamma (t)$ is not rectifiable? I tried to show that $L_{(-1,1)} \ (\gamma(t))\nless \infty$ but i failed.
The length of the curve would be $$\int^1_{-1}\sqrt{1+f'^2(t)}\,dt>\int^1_{-1}|f'(t)|\,dt,$$ but the latter integral diverges.