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. How many new cars should not be added each year to meet this requirement?
(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 an annual increase of $x$ ten thousand cars, 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 $30-\frac{x}{0.06}1.8$, $\lim _{n \rightarrow \infty}\left[\frac{x}{0.06}+\left(30-\frac{x}{0.06}\right) \times 0.94^{\pi-1}\right]=\frac{x}{0.06}$, at this point $\left\{b_{n}\right\}$ 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 annual increase in the number of cars should not exceed 3.6 ten thousand.