Define $\phi(n,x)= \sum\_{m\leq x,\gcd(m,n)=1} 1$, the number of elements in the interval $[1,x]$ that is relatively prime to $n$. $\omega(n)$ is the number of distinct prime factors of $n$.


It's not difficult to show that $\phi(n,x) \geq x \phi(n)/n - 2^{\omega(n)}$ (see [this question](https://mathoverflow.net/questions/88777/bound-the-error-in-estimating-a-relative-totient-function)). This shows for large $x$, say $x\geq n^\frac{1}{\log \log n}$, $\phi(n,x)$ is essentially $x\phi(n)/n$.


Are there better bounds known for smaller values of $x$, say $x=o(2^{\omega(n)})$? The bound is negative for $x$ in those values. 


Even the naive bound $\phi(n,x)\geq 1+\pi(x)-\omega(n)$ performs better when $x$ is small.



Bounds for relative totient function for small values
The fundamental lemma of sieve theory will give you good results here. Put $n\_0 = \prod\_{p|n, p< (\log n)^2} p$, so that $n\_0$ is the product of the small prime factors of $n$. The fundamental lemma gives 
$$ 
\Big| \sum\_{\substack{m \le x\\ (m,n\_0)=1}} 1 - \frac{\phi(n\_0)}{n\_0} x \Big| \ll u^{-u(1+o(1))} x \frac{\phi(n\_0)}{n\_0}, 
$$ 
with $x= (\log n)^{2u}$ (and $u \ge 2$ say). 


Next note that there cannot be too many primes dividing $n$ larger than $(\log n)^2$, and these make a negligible contribution: 
$$ 
\Big| \sum\_{\substack{m\le x \\ (m,n)=1} }1 - \sum\_{\substack{m\le x \\ (m,n\_0)=1} }1 \Big| \le \sum\_{\substack { p|n \\ p>(\log n)^2}} \frac xp = o\Big(\frac{x}{\log n}\Big). 
$$ 


Thus it follows that you can get a good asymptotic formula whenever $x$ is larger than $(\log n)^{2u}$ for any slowly growing quantity $u$. Consult any standard text on analytic number theory (for example Montgomery and Vaughan) for the fundamental lemma. 


