We have the following equation $$1-\frac{h}{u}=\frac{R}{\sqrt{u^2+1}}.$$ Let $r=R-1$ then show that for small $r,h,u$ the equation reduces to $$h+ru-\frac{u^3}{2}\approx0.$$
My Attempt: On squaring both sides we get $$1+\frac{h^2}{u^2}-\frac{2h}{u}=\frac{R^2}{u^2+1}$$ $$\implies \color{red}{u^4}+u^2h^2+h^2-2hu^3-\color{red}{2hu}=(r^2+1+\color{red}{2r})u^2$$ If keep only $$u^4-2hu=2ru^2$$ Then I can get $$ru+h-\frac{u^3}{2}\approx 0.$$ But why does this work?
I think there need to be further assumptions on the relative rates at which $r$ and $h$ approach zero.
We're tasked with developing the solutions for $u$ of the quartic equation $$u^4 -2hu^3 +(h^2-2r-r^2) u^2 -2hu +h^2=0$$ as series in $r$ and $h$. But we don't know the relative rates at which $r$ and $h$ approach zero, so let's suppose that
$$\begin{align}r&\propto t^{\mathrm{ord}\,r}(1+o(1))\\ h&\propto t^{\mathrm{ord}\,h}(1+o(1))\\ u&\propto t^{\mathrm{ord}\,u}(1+o(1))\text{.} \end{align}$$ Then for $u^4$, $-2hu$, and $-2ru^2$ to be the lowest order vanishing terms in the quartic polynomial, calculations involving the Newton polygon imply that we must have $$\begin{align} 0<\mathrm{ord}\,u &=\mathrm{ord}\,h-\mathrm{ord}\,r\\ 0<\mathrm{ord}\,r &<2(\mathrm{ord}\,h)\\ 0<\mathrm{ord}\,h &<2(\mathrm{ord}\,r)\text{,} \end{align}$$ e.g., $$\begin{align}\mathrm{ord}\,r&=2&\mathrm{ord}\,h&=3&\mathrm{ord}\,u&=1\text{.} \end{align}$$