Currently, I'm working on problem 4.21 in Apostol's Analytic number theory and I solved that in different ways. The original problem is as follows:
Given two real-valued functions $S(x)$ and $T(x)$ such that $$T(x)=\sum_{n\le x}S(\frac xn),x\ge1.$$ If $S(x)=O(x)$ and if $c$ is a positive costant, prove that the relation
$$S(x)\sim cx~\text{as}~x\to\infty$$ implies
$$T(x)\sim cx\log x~\text{as}~x \rightarrow \infty.$$
This is one of my answers:
$$\lim_{x\to\infty}\frac{T(x)}{cx\log(x)}=\lim_{x\to\infty}\frac{\sum_{n\le x}S(\frac xn)}{cx\log(x)}=\lim_{x\to\infty}\frac{\sum_{n\le x}(\frac{cx}n+o(x))}{cx\log(x)}=\lim_{x\to\infty}\frac{\sum_{n\le x}\frac 1n}{\log x}=\lim_{x\to\infty}\frac{\log x+O(1)}{\log x}=1$$
As you see, I didn't use the relation $S(x)=O(x)$. But the problem has said that "If $S(x)=O(x)$". Could anyone help me to know what is wrong in my answer? Thanks!
The assumption $S(x)=O(x)$ follows from $S(x)\sim cx$ and therefore is totally redundant.
On the other hand, there is an inaccuracy in your argument: your $o(x)$ term actually depends not only on $x$, but also on $n$ and, more importantly, you cannot dispose of this term the way you did. You should instead define $\phi(x)$ by $S(x)=cx(1+\phi(x))$, so that $\phi(x)\to 0$ and argue more accurately in terms of this function.