Example 4 Let $x>0$, find the range of the function
$$
f(x)=\frac{x+\frac{1}{x}}{[x] \cdot\left[\frac{1}{x}\right]+[x]+\left[\frac{1}{x}\right]+1}
$$

where $[x]$ denotes the greatest integer not exceeding $x$.
Given that the expression of $f(x)$ is symmetric with respect to $x$ and $\frac{1}{x}$, and $x > 0$, we can assume without loss of generality that $x \geqslant 1$.
(i) When $x=1$, $\frac{1}{x}=1$, we have $f(1)=\frac{1}{2}$;
(ii) When $x>1$, let $x=n+\alpha, 0 \leqslant \alpha < 1$, and $b_{1}>b_{2}>b_{3}>b_{4}>\cdots>b_{n}>\cdots$
Thus, $\left[a_{2}, b_{2}\right)=I_{2} \supsetneq I_{3} \ni I_{4} \ni \cdots \ni I_{n} \geqslant \cdots$
Since $I_{1}=\left[a_{1}, b_{1}\right)=\left[1, \frac{5}{4}\right), I_{2}=\left[a_{2}, b_{2}\right)=\left[\frac{5}{6}, \frac{10}{9}\right)$,

Therefore, $I_{1} \cup I_{2} \cup I_{3} \cup \cdots \cup I_{n} \cup \cdots=I_{1} \cup I_{2}=\left[1, \frac{5}{4}\right) \cup\left[\frac{5}{6}, \frac{10}{9}\right)=\left[\frac{5}{6}, \frac{5}{4}\right]$.
In conclusion, the range of $f(x)$ is $\left\{\frac{1}{2}\right\} \cup\left[\frac{5}{6}, \frac{5}{4}\right)$.