Example 2: By the end of 2001, the number of cars in a city was 300,000. It is expected that each year, 6% of the car stock from the previous year will be scrapped, and the number of new cars added each year will be the same. To protect the urban environment, the city's car stock should not exceed 600,000. What is the maximum number of new cars that can be added each year?
(2002 National College Entrance Examination Question)
Let the number of cars at the end of 2001 be $b_{1}$ ten thousand, and the number of cars at the end of each subsequent year be $b_{2}$ ten thousand, $b_{3}$ ten thousand, $\cdots \cdots$, with $x$ ten thousand new cars added each year, then $b_{1}=30, b_{2}=b_{1} \times 0.94+x$.
When $n>1$, we have $b_{n+1}=b_{n} \times 0.94+x=b_{n-1} \times 0.94^{2}+(1+0.94) x=\cdots$
$b_{n+1}=b_{1} \times 0.94^{n}+x\left(1+0.94+\cdots+0.94^{n-1}\right)=\frac{x}{0.06}+\left(30-\frac{x}{0.06}\right) \times 0.94^{n}$.
When $30-\frac{x}{0.06} \geqslant 0$, i.e., $x \leqslant 1.8$, then $b_{n+1} \leqslant b_{n} \leqslant \cdots \leqslant b_{1}=30$.
When $30-\frac{x}{0.06}1.8$, $\lim _{n \rightarrow \infty}\left[\frac{x}{0.06}+\left(30-\frac{x}{0.06}\right) \times 0.94^{-1}\right]=\frac{x}{0.06}$, at this time $\left\{b_{n}\right\}$

is monotonically increasing and can approach $\frac{x}{0.06}$ infinitely. Therefore, $\frac{x}{0.06} \leqslant 60, x \leqslant 3.6$.
In summary, the number of new cars added each year should not exceed 3.6 ten thousand.