Asymptotic Estimate of Vector Function I would like to compute the asymptotic limit of the of the following function
$$f(x,\omega) = \frac{x - \omega\sqrt{1+|x|^2}}{x\cdot \omega - \sqrt{1+|x|^2}}$$
Where $x\in \mathbb{R}^3$ and $\omega \in \mathbb{S}^2 = \{y\in\mathbb{R}^3 \ | \ |y| = 1\}$ is a point on the unit sphere. More precisely, I need to estimate $||f(x,\cdot)||_{L^\infty(\mathbb{S^2)}} := \sup_{\omega\in\mathbb{S}^2}|f(x,\omega)|$ for large $|x|$. I have the rough estimate
\begin{align}
|f(x,\omega)| &\leq \bigg|\frac{\frac{x}{\sqrt{1+|x|^2}} - \omega}{\frac{x\cdot\omega}{\sqrt{1+|x|^2}} - 1}\bigg| \leq \frac{2}{1 - \frac{|x|}{\sqrt{1+|x|^2}}}
\\
&= \frac{2\sqrt{1+|x|^2}}{\sqrt{1+|x|^2} - |x|} = \mathcal{O}(|x|^2)
\end{align}
as $|x|\rightarrow \infty$. But I am wondering if I can do better than this and obtain a sharper estimate (possibly $\mathcal{O}(|x|)$ or even $\mathcal{O}(1)$)?
Edit:
To elaborate I am looking for a better function $g(|x|)$ such that
\begin{align}
\sup_{(\hat{x},\omega)\in\mathbb{S}^2\times\mathbb{S}^2}|f(|x|\hat{x},\omega)| \leq C g(|x|)
\end{align}
I suspect something like linear in $|x|$ like $g(|x|) = a|x| + b$.

For $x \to \infty$, there is very good approximation
$$f(x,\omega) = \frac{x - \omega\sqrt{1+|x|^2}}{x\, \omega - \sqrt{1+|x|^2}}=-1-\frac{\omega +1}{2 x^2 (\omega -1)}+\frac{\omega ^2-2 \omega -3}{8 x^4 (\omega
   +1)^2}+O\left(\frac{1}{x^6}\right)$$ At the opposite, for $x \to -\infty$, I did not find anything even decent.
