Interesting solution. Mine based on yours... :-)

1) f(n) > (n/1+n/2+...+n/n)/n-1=1/2+1/3+...+1/n, so the limit of f(n) is +infinity

2) Let n+1 be a prime... f(n+1)=([(n+1)/1]+...+[(n+1)/n]+[(n+1)/(n+1)])/(n+1)=(1+[n/1]+[n/2]+...+[n/n]+1)/(n+1)=(2+nf(n))/(n+1) < f(n) when 2 < f(n).