Firstly you're wrong when you comment that bijection implies ratio is 1:1. A quick example is the two sets A = {0, 1, 2, 3...} and B = {0, 2, 4, 6..}. These sets form a bijection (a -> 2a, and b -> (1/2)b). The ratio of #s in B less than x over #s in A less than x for some x is (ceiling(x/2) / x). And so lim x-> inf of (ceiling(x/2) / x) = 1/2. So bijections don't imply ratio is 1:1.

Secondly, you're wrong about their example not working out. The ratio in their example does not become 1:1, the limit tends to 0. Quick explanation, simpler ways exist, but just for illustration:

From the prime number theorem, we know lim x-> inf of (pi(x)/(x / (log x)) = 1, where pi(x) counts the number of primes below x.

In this case we want to figure out what limit of the ratio of primes is to non primes below x, or lim x->inf of (pi(x)/(x - pi(x)). Dividing both numerator and denominator by (x/(log(x)), we have lim x-> inf of pi(x)/(x/(log(x)) / ((x - pi(x))/ (x / log(x)).

Quotient limit rule, so the numerator limit is 1 by prime number theorem, so we have it equivalent to 1 / (lim x-> inf of ((x - pi(x)) / (x / log(x))). So if we show the denominator limit goes to infinity, the entire limit is 0.

Bottom limit simplifies to lim x-> inf of x*log(x)/x - log(x) * pi(x)/x, the latter term goes to 1 by the prime number theorem, and thus the entire denominator limit is just lim x-> inf of (log(x) - 1) which goes to infinity, thus entire limit goes to 0.