jmc

Rationalizing the numerator, we get

$$\begin{array}{rl}\frac{x^2+2-\sqrt{x^4+4}}{x}\cdot \frac{x^2+2+\sqrt{x^4+4}}{x^2+2+\sqrt{x^4+4}}& =\frac{(x^2+2{)}^2-(x^4+4)}{x(x^2+2+\sqrt{x^4+4})}\\ & =\frac{4x^2}{x(x^2+2+\sqrt{x^4+4})}\\ & =\frac{4}{\frac{1}{x}(x^2+2+\sqrt{x^4+4})}\\ & =\frac{4}{x+\frac{2}{x}+\sqrt{x^2+\frac{4}{x^2}}}.\end{array}$$Since we wish to maximize this quantity, we wish to minimize the denominator. By AM-GM, $x+\frac{2}{x}\ge 2\sqrt{2}$ and $x^2+\frac{4}{x^2}\ge 4$, so that the denominator is at least $2\sqrt{2}+2$. Therefore, $$\frac{x^2+2-\sqrt{x^4+4}}{x}\le \frac{4}{2\sqrt{2}+2}=\boxed{2\sqrt{2}-2},$$with equality when $x=\sqrt{2}$.