Consider $T, g: \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ which are bounded on any bounded interval, satisfy $$T(x)=2 T(x / 2)+g(x)$$ for every real number $x \geqslant 1$ and $g(x)=\mathcal{O}(x)$.
I would like to show $T(x)=\mathcal{O}(x \log (x))$. For this I want to show $\lim_{x\rightarrow \infty} \dfrac{T(x)}{x\log(x)}$ is finite.
Notice $$T(x)=2^nT(x/2^n)+2^{n-1}g(x/2^{n-1})+\ldots+2g(x/2)+g(x)\leq2^nT(x/2^n)+nCx$$ where $C$ is a constant such that $g(x)\leq Cx$.
Then $$\lim\limits_{x\rightarrow \infty} \dfrac{T(x)}{x\log(x)}\leq \lim\limits_{x\rightarrow \infty} \dfrac{2^nT(x/2^n)+nCx}{x\log(x)}=\lim\limits_{x\rightarrow \infty} \dfrac{2^nT(x/2^n)}{x\log(x)}.$$
Now I tried taking $n\rightarrow \infty$ and setting $x=2^n$ to simplify all terms but I am really not sure I am allowed to do this.
Do you have any idea how to finish the proof?
You don't need to prove that the limit is finite the limit can even not exist. However, you can easily prove that the function is bounded. Indeed, for $x\in \mathbb R_{>0}$, $n = \left\lceil\log_2(x)\right\rceil$ then $$\log_2(x) \le n \le \log_2(x) + 1 \implies x \le 2^n \le 2x \implies \frac12 \le \frac{x}{2^n} \le 1$$ and
\begin{align} \frac{T(x)}{x\log_2(x)} &\le \frac{T\left(\frac x{2^n}\right)}{\frac x{2^n} \log_2 x} + C\frac{n}{\log_2x}\\ &= \frac{\sup_{t\in\left[\frac12,1\right]} T(t)}{2 \log_2(x)} + C\left(1 + \frac1{\log_2(x)}\right) \end{align}
Can you finish the proof?