The cutoff phenomenon for Markov chains is defined as follows
Suppose for a sequence of Markov Chains indexed by $n = 1, 2, \cdots$, the mixing time for the $n$-th chain at kevel $\epsilon$ is denoted by $t_{mix}^{n}(\epsilon)$. The sequence of chains has a $\textbf{cutoff}$ if for all $\epsilon > 0$ , $$\lim_{n\to\infty} \displaystyle\frac{t_{mix}^{n}(\epsilon)}{t_{mix}^{n}(1 - \epsilon)} = 1$$I came across a equivalent definition which says that the sequence has a cutoff if the total variation distance from the stationary distribution $d_n(t)$ satisfies $$d_n((1-\epsilon)t_{mix}^n(1/2)) \to 1 \\d_n((1+\epsilon)t_{mix}^n(1/2)) \to 0 $$ How are these two definitions equivalent?