Problem has to be solved specifically using little o function. I was going to transform $\sqrt{x+1}$ into $1+\frac{1}{2}x+o(x)$ and $\sqrt{2x}$ into $1+\frac{1}{2}t+o(t)=1+\frac{1}{2}(2x-1)+o(2x)$ but I don't know what to do with $\sqrt{x-1}$. Then, I tried:
$$\lim_{x\to\infty} (\sqrt{x+1} + \sqrt{x-1}-\sqrt{2x})=\lim_{x\to\infty} (1+\frac{1}{2}x+o(x)-1-\frac{1}{2}(2x-1)+o(2x) +\sqrt{x-1})=\lim_{x\to\infty}(\frac{1}{2}x+o(x)-\frac{1}{2}(2x-1)+o(2x) +\sqrt{x-1})$$
What should I do next (if this is correct)? What to do with two different o-functions ($o(x)$ and $o(2x)$) and with $\sqrt{x-1}?$
Your expansion $\sqrt{x+1}=1+\frac12x+o(x)$ is only valid for small $x$, where $o(x)$ is understood as applying when $x\to 0$. When $x\to\infty$, you need $$\sqrt{x+1}=\sqrt x\left(1+\frac1x\right)^\frac12=\sqrt x\left(1 + \frac{1}{2x}+o\left(\frac1x\right)\right)=\sqrt x+\frac{1}{2\sqrt x}+o(x^{-3/2})$$
It is true that $o(x)$ is ambiguous in this regard: you must rely on context to deduce whether it applies as $x\to 0$ or as $x\to\infty$.