Consider the space curve $c : (-\frac{1}{2},\frac{1}{2}) \rightarrow \mathbb{R^3}$ , $$c(x)= \begin{cases} (x,e^{-\frac{1}{x^2}},0) &\ x<0\\ \ \\ (0,0,0)&\ x=0 \\ \ \\ (x,0,e^{-\frac{1}{x^2}}) &\ x>0 \end{cases} $$
Show:
$\cdot$ $c$ is a regular $C^{\infty}$ curve and $ \kappa: (-\frac{1}{2},\frac{1}{2}) $ $\rightarrow \mathbb{R_{\geq 0}}$ a $C^{\infty}$ function.
$\cdot$ $B(x) =(0,0,1)$ for $x < 0$ and $B(x) = (0,-1,0)$ for $x>0$ .
$\kappa$ = curvature with $\kappa(x)$ := $\frac{|| c' \times c''||(x)}{||c'(x)||^3} $.
$B$ = binormal with $B(x)$ := $T(x) \times N(x)$
$\tau$ = torsion with $\tau(x)$ := $\frac{det(c',c'',c''')(x)}{|| c' \times c''||^2(x)} $
$T(x)$ := $\frac{1}{||c'(x)||} \cdot c'(x) $.
$N(x)$ := $\frac{1}{||T'(x)||} \cdot T'(x) $.
My thoughts: a curve is regular, if $|| c'(x) || \neq 0$. $$c'(x)= \begin{cases} (1,\frac{2}{x^3}e^{-\frac{1}{x^2}},0) &\ x<0\\ \ \\ (1,0,0)&\ x=0 \\ \ \\ (1,0,\frac{2}{x^3}e^{-\frac{1}{x^2}}) &\ x>0 \end{cases} $$ Do I have a mistake here? Next point: It's clear to me that $c$ is a $C^{\infty}$ function for $x \neq 0$, but I have problems with $x=0$. Next point: for the curvature we need $c''$.$$c''(x)= \begin{cases} (0,-\frac{(6x^2-4)}{x^6}e^{-\frac{1}{x^2}},0) &\ x<0\\ \ \\ (0,0,0)&\ x=0 \\ \ \\ (0,0,-\frac{(6x^2-4)}{x^6}e^{-\frac{1}{x^2}}) &\ x>0 \end{cases} $$
$$\kappa(x) \begin{cases} \frac{|\frac{(6x^2-4)}{x^6}e^{-\frac{1}{x^2}}|}{(1+(\frac{2}{x^3} e^{-\frac{1}{x^2}})^2)^{\frac{3}{2}}} &\ x<0\\ \ \\ 0&\ x=0 \\ \ \\ \frac{|\frac{(6x^2-4)}{x^6}e^{-\frac{1}{x^2}}|}{(1+(\frac{2}{x^3} e^{-\frac{1}{x^2}})^2)^{\frac{3}{2}}} &\ x>0 \end{cases} $$
I know that my result looks terrible. I really need help here. Is there a trick? How can I solve the other claims? I mean how can I solve them, if my results looks so terrible.