We can look at this question as asking whether the product of (p-1)/p for all primes p goes to zero. Taking the log, this is the same as asking whether the sum

We can look at this question as asking whether the product of (p-1)/p for all primes p goes to zero. Taking the log, this is the same as asking whether the sum of log((p-1)/p) is unbounded. log((p-1)/p) = log(p-1) - log(p), which is less than -1/p (greater in magnitude). The sum of the reciprocals of the primes is unbounded (https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_r...). So the sum of the logs is less than the negative of the sum of the reciprocals of the primes, it is unbounded as well. Therefore the limit is 100.