I am trying to prove that $v \in T_{S}(x)$ if and only if $\lim \inf_{t \downarrow 0}d_{S}(x+tv)/t=0$ and $d_S$ is the distance function of $S$ and S is a subset of the normed space $X$.
For the necessity, $v \in T_{S}(x)$, then by the definition of tangent cone, there is a sequence $v_i$ in X converging to $v$, and a positive sequence $t_i$ decreasing to $0$, such that $x+t_iv_i \in S \ \forall i$. Since $d_{S}(x+tv) \geq 0$, then we have $\lim \inf_{t \downarrow 0}d_{S}(x+tv)/t \geq 0$, on the other hand, $\lim \inf_{t \downarrow 0}d_{S}(x+tv)/t \leq \lim_{i \to \infty}d_{S}(x+t_{i}v_{i})/t_i=0$, deduce that $\lim \inf_{t \downarrow 0}d_{S}(x+tv)/t=0$. I am not sure the proof is correct and fail to prove the opposite direction. Any help will appreciated.
I think there's a missing step before your inequality $\liminf_{t\downarrow 0}d_S(x+tv)/t\leqslant \lim_{i\rightarrow \infty}d_S(x+t_iv_i)/t_i$ for the necessity. It is the fact that $d_S(x+t_iv)\leqslant d_S(x+t_iv_i)+t_i||v_i-v||$ thanks to $d_S$ being 1-lipschitzian.
For the opposite direction, there exists sequences $t_i>0$ decreasing to $0$ and $\varepsilon_i>0$ converging to $0$ such that $d_S(x+t_iv)<\varepsilon_i t_i$ hence there exists a sequence $x_i\in S$ such that $||x+t_iv-x_i||<2\varepsilon_i t_i$ $(*)$. Put $v_i=\frac{x_i-x}{t_i}$ : $x_i=x+t_i v_i$ so from $(*)$ we get $||v-v_i||<2\varepsilon_i$ and thus $v_i$ converges to $v$, $x+t_i v_i$ belongs to $S$ so $v\in T_S(x)$.