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 $g(|x|) = a|x| + b$.
 A: Let $x=|x|\hat{x}$ and $\hat{x}=(\hat{x}\cdot\omega)\omega+\alpha\omega^\perp$, then
$$
f(x,\omega)=\frac{x-\omega\sqrt{1+|x|^2}}{x\cdot\omega-\sqrt{1+|x|^2}}=\omega+\frac{\alpha}{\hat{x}\cdot\omega-\sqrt{(1+|x|^2)/|x|^2}}\omega^\perp$$
Hence for  $\xi=\sqrt{(1+|x|^2)/|x|^2}$, \begin{align*}|f(x,\omega)|^2&=1+\frac{\sin^2\theta}{(\cos\theta-\xi)^2}\\
&=1+\sin^2\theta(\xi^2-2\xi\cos\theta+\cos^2\theta)^{-1}\\
&=1+\sin^2\theta\left((1-\cos\theta)^2+\frac{1-\cos\theta}{|x|^2}+O(|x|^{-4})\right)^{-1}\\
&=1+\frac{\sin^2\theta}{(1-\cos\theta)^2}\left(1-\frac{1}{(1-\cos\theta)|x|^2}+O(|x|^{-4}) \right)
\end{align*}
Finally taking the square root, $$\fbox{$|f(x,\omega)|=\frac{1}{\sin(\theta/2)} - \frac{\sin^2\theta}{16\sin^5(\theta/2)}\frac{1}{|x|^2}+O(|x|^{-4})$}$$
Note: $\lim_{|x|\to\infty}|f(x,\omega)|=1/\sin(\theta/2)$, which has no sup near $\theta\to0$.

Edit: At $\theta=0$, i.e., $\hat{x}=\omega$, $f(x,\omega)=1$ exactly, but for nearby values, $f$ is unbounded.
Edit2: A plot of $|f(x,\omega)|$ and the approximation agree.
A: 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.
