The Chebyshev functions are defined as $\psi(x) = \sum_{p^m \leq x} \log n$ and $\theta(x) = \sum_{p\leq x} \log p$, where $p$ is a prime, $m\geq 1$ is an integer and $n=p^m$ in $\psi(x)$. It is known that there exist positive constants $c_{1}$ and $c_2$ such that $c_{1}x < \psi(x) < c_{2} x$ and $ \frac{1}{2}c_{1}x < \theta(x) < c_{2} x$. By these bounds we find that $ \dfrac{ \psi(x) - x}{\psi(x) - \theta(x)} < \dfrac{(c_{2} -1)x}{(c_2 - c_1)x} = \dfrac{c_{2} -1 }{c_2 - c_1} <\infty $.
Hence invoking the well known result $\mid \psi(x) - \theta(x) \mid = O(x^{\frac{1}{2}}\log x)$, it then follows that $\mid \psi(x) - x \mid = O(x^{\frac{1}{2}}\log x)$ ?
EDIT: It is also known that $\theta (x)$ tends to $x$ as $x$ tends to $\infty$. Surely, an impication of this is $\dfrac{ \psi(x) - x}{\psi(x) - \theta(x)} < \infty$ since $\psi(x) \neq \theta(x)$.