The other replies are good, but let's add another one anyway.

0.987654321/0.123456789 = (1.11111111-x)/x = 1.11111111/x - 1 where x = 0.123456789

You can aproxímate 1.11111111 by 10/9 and aproxímate x = 0.123456789 using y = 0.123456789ABCD... = 0.123456789(10)(11)(12)(13)... that is a number in base 10 that is not written correctly and has digits that are greater than 9. I.E. y = sum_i>0 i/10^i

Now you can consider the function f(t) = t + 2 t^2 + 3 t^3 + 4 t^4 + ... = sum_i>0 i*t^i and y is just y=f(0.1).

And also consider an auxiliary function g(t) = t + t^2 + t^3 + t^4 + ... = sum_i>0 1*t^i . A nice property is that g(t)= 1/(1-t) when -1<t<1.

The problem with g is that it lacks the coefficients, but that can be solved taking the derivative. g'(t) = 1 + 2 t + 3 t^2 + 4 t^3 + ... Now the coefficients are shifted but it can be solved multiplying by t. So f(t)=t*g'(t).

So f(t) = t * (1/(1-t))' = t * (1/(1-t)^2) = t/(1-t)^2

and y = f(0.1) = .1/.9^2 = 10/81

then 0.987654321/0.123456789 ~= (10/9-y)/y = 10/(9y)-1 = 9 - 1 = 8

Now add some error bounds using the Taylor method to get the difference between x and y, and also a bound for the difference between 1.11111111 an 10/9. It shoud take like 15 minutes to get all the details right, but I'm too lazy.

(As I said in another comment, all these series have a good convergence for |z|<1, so by standards methods of complex analysis all the series tricks are correct.)