Solution:
$2 \sqrt{2}-2$

Rationalizing the numerator, we get
$$\begin{aligned} \frac{x^{2}+2-\sqrt{x^{4}+4}}{x} \cdot \frac{x^{2}+2+\sqrt{x^{4}+4}}{x^{2}+2+\sqrt{x^{4}+4}} & =\frac{\left(x^{2}+2\right)^{2}-\left(x^{4}+4\right)}{x\left(x^{2}+2+\sqrt{x^{4}+4}\right)} \\ & =\frac{4 x^{2}}{x\left(x^{2}+2+\sqrt{x^{4}+4}\right)} \\ & =\frac{4}{\frac{1}{x}\left(x^{2}+2+\sqrt{x^{4}+4}\right)} \\ & =\frac{4}{x+\frac{2}{x}+\sqrt{x^{2}+\frac{4}{x^{2}}}} \end{aligned}$$
Since we wish to maximize this quantity, we wish to minimize the denominator. By AM-GM, $x+\frac{2}{x} \geq 2 \sqrt{2}$ and $x^{2}+\frac{4}{x^{2}} \geq 4$, so that the denominator is at least $2 \sqrt{2}+2$. Therefore,
$$\frac{x^{2}+2-\sqrt{x^{4}+4}}{x} \leq \frac{4}{2 \sqrt{2}+2}=2 \sqrt{2}-2,$$
with equality when $x=\sqrt{2}$.