I was reading through these notes (link here) on Markov diffusion processes and I'm confused about one of the definitions and its properties.
It is stated that the Markov diffusion process satisfies the continuity condition in that $P(|X_t - X_s | \ge \epsilon | X_s = x) = o(t-s)$. It then goes on to say that this condition implies that the sample paths of a diffusion process are not differentiable.
I am confused by this statement. From what I understand, the $o$-notation here means that $\lim_{t\rightarrow s} \frac{o(t-s)}{t-s} = 0$. For example, we have that $x^5$ is $o(x)$ as $x\rightarrow 0$. This makes me think that if $f(x)$ is $o(x)$ as $x \rightarrow 0$, i.. the rate of decay of $f(x)$ compared to that of $x$ is faster as $x \rightarrow 0$.
Hence, I do not understand why in the link it says: "the sample paths of a diffusion process are not differentiable : if they where, then the right hand side of the above equation would have to be 0 when $t-s \ll 1$ ".
Can anyone help me show (more rigorously) how this implies that $P(\frac{|X_t - X_s |}{t-s} \ge \epsilon | X_s = x) $ does not approach 0 as $t \rightarrow s$?