Consider a smooth plane curve $c:I\to \mathbb{R}^2$ with $I$ an open interval in the real line and $c' \neq 0$ everywhere. Assume $t_0 \in I$. My question is whether following statements are equivalent:
(1) $c''(t_0) \neq 0$
(2) There is a sequence $(h_n)_{n \in \mathbb{N}}$ with $h_n \to 0$ and all $h_n>0$ such that for each $n$, the three points $c(t_0 \pm h_n)$, $c(t_0)$ are not located on a straight line.
If this is true, what is a proof? If it is false, what is a counterexample? And if it is false, is there any modifiaction of statement (1) to make the equivalence true? Or a modification of statement (2) or both?
Edit: Thanks to the answers and comments, I learned that (1) $\Rightarrow$ (2) is false without further assumptions. But what about the direction (2) $\Rightarrow$ (1)?
As Eric noted in the comments we can find a counterexample. It can be even as simple as: $$c(t) = \left(t^2,0\right).$$ Obviously, the curve $c$ is flat, so you can choose an arbitrary $t_0 \in \mathbb{R}$ and any real sequence $(h_n)_{n \in \mathbb{N}}$ convergent to $0$. On the other hand, we have $c''(t) = (2,0)$.
The key notion is how fast we move along the line. To make your statements equivalent you can add an assumption that the curve $c$ has to be parameterized by arc length i.e. $\lVert c'(t) \rVert = 1$.