Example 5. If $x \neq 0$, then the maximum value of $\frac{\sqrt{1+x^{2}+x^{4}}-\sqrt{1+x^{4}}}{x}$ is $\qquad$ . (1992, National Junior High School Competition)
Solution 1 $\frac{\sqrt{1+x^{2}+x^{4}}-\sqrt{1+x^{4}}}{x}$
$$
\begin{array}{l}
=\frac{x}{\sqrt{1+x^{2}+x^{4}}+\sqrt{1+x^{4}}} \\
=\frac{x}{|x|\left(\sqrt{x^{2}+\frac{1}{x^{2}}+1}+\sqrt{x^{2}+\frac{1}{x^{2}}}\right)} \\
=\frac{x}{|x|\left(\sqrt{\left.x-\frac{1}{x}\right)^{2}+3}+\sqrt{\left.\left(x-\frac{1}{x}\right)^{2}+2\right)}\right.} .
\end{array}
$$

Therefore, when $x=\frac{1}{x}>0$, i.e., $x=1$, the original expression reaches its maximum value $\sqrt{3}-\sqrt{2}$.
Solution 2 When $x>0$,
$$
\begin{array}{l}
\frac{\sqrt{1+x^{2}+x^{4}}-\sqrt{1+x^{4}}}{x} \\
=\sqrt{\left(x-\frac{1}{x}-0\right)^{2}+(0-\sqrt{3})^{2}} \\
-\sqrt{\left(x-\frac{1}{x} \cdots 0\right)^{2}+(0-\sqrt{2})^{2}} .
\end{array}
$$

Let $t=x-\frac{1}{x}$, then
$$
\begin{aligned}
\text { Original expression }= & \sqrt{(t-0)^{2}+(0-\sqrt{3})^{2}} \\
& -\sqrt{(t-0)^{2}+(0-\sqrt{2})^{2}} .
\end{aligned}
$$

By the distance formula between two points, the geometric meaning of the above expression is: the difference in distances from a moving point $P(t, 0)$ to two fixed points $A(0, \sqrt{3}), B(0, \sqrt{2})$ (as shown in the figure).

Since $A B \geqslant P A-P B$, and $A B=\sqrt{3}-$ $\sqrt{2}$, the point $P$ is at the origin, i.e., $t=0$, when $P A-P B$ reaches its maximum value $\sqrt{3}-\sqrt{2}$.

And when $t \doteq 0$, $x-\frac{1}{x}=0$, we get $x=1$, i.e., when $x=1$, the original expression reaches its maximum value $\sqrt{3}-\sqrt{2}$.