Let $f(x) = x^{2} + \dfrac{1}{x^{2}}$ and $g(x) = x - \dfrac{1}{x}$, $x \in \mathbb{R} - {-1,0,1}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value of $h(x)$ is