Everyone knows (but no one can prove) that there are infinitely many primes $p$ such that $q=2p-1$ is also prime. $\sigma(q)=q+1=2p$, $\phi(\sigma(q))=\phi(2p)=p-1$, $\phi(\sigma(q))/q=(p-1)/(2p-1)\to1/2$ as $q\to\infty$.
Edit: I don't know why it didn't occur to me to look at Guy, Unsolved Problems In Number Theory. Under B42, he writes, "Makowski and Schinzel prove that $\limsup\phi(\sigma(n))/n=\infty$. The reference is A. Makowski, A. Schinzel, On the functions $\phi(n)$ and $\sigma(n)$, Colloq Math 13 (1964-65) 95-99, MR 30 #3870. I haven't found the paper on the web, but it's in Volume 2 of Schinzel's Selecta, 890-894. I don't have the energy to write out the proof in full, but here's the idea. Given $M$, choose $t$ such that $$\prod_{i=1}^t{p_i\over p_i-1}>M$$. Then given $p$ (and it's not clear to me whether $p$ is meant to be a prime), and letting $$n=\sigma\left(\prod_{i=1}^tp_i^{p-1}\right),$$ we get $$\sigma(n)=\prod_{i=1}^tN(p_i,p),$$ where $N(a,p)=(a^p-1)/(a-1)$. Now you prove $\limsup_{p\to\infty}\phi(\sigma(n))/n\gt M$, using along the way a lemma which says that $$\lim_{p\to\infty}{\phi(N(a,p))\over N(a,p)}=1.$$
Everyone knows (but no one can prove) that there are infinitely many primes $p$ such that $q=2p-1$ is also prime. $\sigma(q)=q+1=2p$, $\phi(\sigma(q))=\phi(2p)=p-1$, $\phi(\sigma(q))/q=(p-1)/(2p-1)\to1/2$ as $q\to\infty$.