Limit $lim_{xrightarrowinfty}sqrt{log(x)log(ax)}-log(x)=frac{log{a}}{2}$

I feel really dumb asking this, but I am stuck. Why is

$$\lim_{x\rightarrow\infty}\sqrt{\log(x)\log(ax)}-\log(x)=\frac{\log{a}}{2}$$

(according to Wolfram Mathematica).

Let's assume that $a>1$. I expand the $\log(ax)$ term and obtain:

$$\lim_{x\rightarrow\infty}\sqrt{\log^2(x)+\log(x)\log(a)}-\log(x)$$

The term inside the square root is larger than $\log^2(x)$ due to addition of $\log(x)$ times a constant, and my intuition suggests that the limit diverges. What am I missing?

Answer

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\begin{align}
&\root{\ln\pars{x}\ln\pars{ax}} - \ln\pars{x}
={\ln\pars{x}\ln\pars{ax} - \ln^{2}\pars{x}
\over \root{\ln\pars{x}\ln\pars{ax}} + \ln\pars{x}}
={\ln\pars{x}\ln\pars{a}
\over \root{\ln\pars{a}\ln\pars{x} + \ln^{2}\pars{x}} + \ln\pars{x}}
={\ln\pars{a}
\over \root{\ln\pars{a}/\ln\pars{x} + 1} + 1}
\stackrel{x \to \infty}{\to} \color{#0000ff}{\Large\half\,\ln\pars{a}}
\end{align}