[8].
Consider pairs of numbers $(9, 1)$, $(98, 12)$, ..., $(98765432, 12345678)$, which are formed by pairing the first number of the numerator with the last number of the denominator, second number of the numerator with the next-to-the last number of the denominator and so on and the last number of the numerator with the first number of the denominator. For each of these pairs, it is easy to check the first entry is at least $8$ times bigger than the second entry. Therefore, the given fraction is at least $8$. On the other hand, except for the first pair, for which $9 = 9 \times 1$, the first entry is less than $9$ times the second entry for all other pairs. Therefore, the fraction is less than $9$ and the answer to the problem is $8$.