I have problem with this exercice from Kristopher Tapp's book "Differential Geometry of Curves and Surfaces"
EXERCISE 1.42. (Page 32) Let $\gamma: I \rightarrow \mathbb{R}^{n}$ be a unit-speed curve. Let $t_{0} \in I$ and assume $\kappa\left(t_{0}\right) \neq 0$. For sufficiently small $h>0$, prove that the three points $\gamma\left(t_{0}-h\right), \gamma\left(t_{0}\right)$ and $\gamma\left(t_{0}+h\right)$ are not collinear.
My attempt: I think the question means: prove that there exists $\epsilon >0$ such that for all $h >0$, if $h<\epsilon$ then the three points $\gamma\left(t_{0}-h\right), \gamma\left(t_{0}\right)$ and $\gamma\left(t_{0}+h\right)$ are not collinear.
I tried proof by contradiction. Assume that there exists a sequence $(h_{n})_{n\geq1}\subset \mathbb{R}^{*}_{+}$ such that $h_{n} \rightarrow 0 $ , and $(\alpha_{n})_{n\geq1}\subset \mathbb{R}$ such that $ \gamma\left(t_{0}+h_{n}\right)-\gamma\left(t_{0}\right)=\alpha_{n}\left( \gamma\left(t_{0}-h_{n}\right)-\gamma\left(t_{0}\right) \right)$
(It's easy to show that $\alpha_{n} \rightarrow-1 $ when $n \rightarrow \infty $),
using Taylor expansion we get
$\left\{\begin{array}{l} \left.\gamma\left(t_{0}+h_{n}\right)-\gamma\left(t_{0}\right)=h_{n} \gamma^{\prime}\left(t_{0}\right)+\frac{h_{n}^{2}}{2} \gamma^{\prime \prime}\left(t_{0}\right)+o\left(h_{n}^2\right)\right) \\ \gamma\left(t_{0}-h_{n}\right)-\gamma\left(t_{0}\right)=-h_{n} \gamma^{\prime}\left(t_{0}\right)+\frac{h_{n}^{2}}{2} \gamma^{\prime \prime}\left(t_{0}\right)+o\left(h_{n}^2\right)\end{array}\right.$
from this and the above equation we get
$\left(1+\alpha_{n}\right) h_{n} \gamma^{\prime}\left(t_{0}\right)+\left(1-\alpha_{n}\right) \frac{h_{n}^{2}}{2} \gamma^{\prime\prime}\left(t_{0}\right)=o\left(h_{n}^{2}\right).$
I can't find any contradiction and I can't find where I can use $\kappa\left(t_{0}\right) \neq 0$.
You have it. You have already agreed that $\alpha_n\to -1$ as $n\to\infty$. So we're going to conclude that the only way your expression can be $o(h_n^2)$ is for $\gamma''(t_0)$ to be $0$. This is equivalent to saying $\kappa(t_0)=0$. (Recall that $\gamma'(t_0)$ and $\gamma''(t_0)$ are orthogonal!)
I think your argument is overly complicated. Rather than introducing the contradiction, the sequence, and the $\alpha_n$, just consider $$\big(\gamma(t_0+h)-\gamma(t_0)\big)\times\big(\gamma(t_0-h)-\gamma(t_0)\big).$$ I would recommend writing $\gamma'(t_0)=T_0$, $\gamma''(t_0)=\kappa_0 N_0$. Now use your Taylor
expression and you'll get $$(h T_0+\tfrac{h^2}2\kappa_0 N_0+o(h^2))\times ({-}h T_0+\tfrac{h^2}2\kappa_0 N_0+o(h^2)) = h^3\kappa_0 B_0 + o(h^3).$$ Since $\kappa_0\ne 0$, the two vectors are indeed linearly independent for $h\ne 0$.