Directional subderivatives: definition and computation For a continuous vector field $f$ and some vector $v$ in a Banach space, the (lower) directional subderivative can be defiend as follows:
$$
 Df_v(x) \triangleq \liminf_{ \substack{ t \rightarrow 0 \\ u \rightarrow v } } \frac{1}{t} ( f(x + t u) - f(x) ), 
$$
What exactly does the $\liminf$ mean in this case?
I understand how to decipher $\liminf_{t \rightarrow 0}$, but what does $u \rightarrow v$ change in the construction?
Also, what is the best way to actually compute the subderivative?

If $(Y,d)$ is a metric space, $E\subseteq Y$, and $h:E\rightarrow
\mathbb{R}$. If $y_{0}\in X$ is an accumulation point of $E$, the liminf of $h$ as $y\to y_0$ is defined as
$$\liminf_{y\rightarrow y_{0}}h(x):=\lim_{r\rightarrow0^{+}}\inf_{E\cap(B(y_{0},r)\setminus\{y_{0}\})}h.$$
Note that this makes sense because if for
every $r>0$ we define
$$
g(r):=\inf_{E\cap(B(y_{0},r)\setminus\{y_{0}\})}h,
$$ then $g(r_{1})\geq
g(r_{2})$ if $r_{1}<r_{2}$. Hence the function $g$ is decreasing and so it has a limit as $r\to 0^+$. 
So in your case you have $Y=\mathbb{R}\times V$ and $y_0=(0,v)$ and $h(t,u)=\frac{f(x+tu)-f(x)}t$.
Hence you are considering
$$\lim_{r\rightarrow0^{+}}\inf_{0<|t|<r,\,0<\Vert u-v\Vert<r}\frac{f(x+tu)-f(x)}t$$
