Showing $\tau(n)/\phi(n)\to 0$ as $n\to \infty$ I was wondering how to show that $\tau(n)/\phi(n)\to 0$, as $n\to \infty$. Here $\tau(n)$ denotes the number of positive divisors of n, and $\phi(n)$ is Euler's phi function.

There is methodology for dealing separately with numerator and denominator, initiated by Ramanujan. The construction of "superior highly composite" numbers gives:
$$ \tau(n) \; \leq \; \; 24 \; \; \left( \frac{n}{315} \right)^{(1/3)} $$ 
The related argument, with a certain type of optima at the primorials, gives
$$ \phi(n) \geq \frac{2 n^{2/3}}{6^{2/3}},$$
and together we get 
$$ \frac{\tau(n)}{\phi(n)} \leq \frac{12 \; \cdot \; 4^{1/3}}{ 35^{1/3} \; \;  n^{1/3}} \; = \; \frac{5.823426...}{n^{1/3}}     $$ 
EDIT, Tuesday, 20 December: very little machinery is required to describe how to get very small values of $\phi(n)$ relative to the size of $n,$ then derive useful inequalities from that information. Given some $1 \geq \delta > 0,$ we wish to find the number $N_\delta$ that gives the smallest value of
$$  f(n) = \frac{\phi(n)}{n^{1-\delta}}.$$ We see that $f(1) = 1.$ Furthermore, if we have some $n$ that is not divisible by a prime $p,$ then $f(np) = f(n) f(p),$ and we shrink the result if we multiply by $p$ when $f(p) < 1.$ At the same time, if $n$ is already divisible by $p,$ multiplying by $p$ always increases $f,$ so $N_\delta$ will be squarefree in any case. So that is the recipe,  $N_\delta$ is the (finite) product of the primes $p$ for which $$ p^{1-\delta} \geq p-1.$$  The derivative of $x^{1-\delta} - x + 1$ is
$(1-\delta) x^{-\delta} -1,$ and this is negative for $x \geq 1,$ while the second derivative $-\delta (1-\delta) x^{-1-\delta}$ is also negative there, so at some point $x^{1-\delta} - x + 1$ becomes $0$ and then negative thereafter. As a result, $N_\delta$ is the product of the consecutive primes up to some point. These are called primorials.
Meanwhile the first (largest) $\delta$ that  includes a prime $p$ is
$$  \delta_p = \frac{\log \left( 1 + \frac{1}{p-1} \right)}{\log p}    $$
Taking $\delta = 1/3,$ we get $N_{1/3}=6$ and $1 -\delta = 2/3,$ so we always have 
$$ \frac{\phi(n)}{n^{2/3}} \geq  \frac{\phi(6)}{6^{2/3}} =  \frac{2}{6^{2/3}}  $$
