Answer
Decrease, $0$.
Work Step by Step
Step 1. Given the equation $L=L_0 \sqrt {1-v^2/c^2}$, as $v$ increases, the term $v^2/c^2$ will also increase; however, the term $1-v^2/c^2$ will decrease, resulting in the decrease of $L$.
Step 2. $\lim_{v\to c^-}L= L_0 \sqrt {1-c^2/c^2}=0$, the left-hand limit is needed because the speed of light is the maximum speed achievable by an object; in other words $v\leq c$.
