Limit example in Zangwill "Modern Electrodynamics" Zangwill shows that the potential of a finite line segment going from $-L$ to $L$ on the $z$-axis with constant line charge density $\lambda$ is:
$$\phi(z,\rho) = \frac{\lambda}{4\pi\epsilon_0}\ln\left(\frac{\sqrt{(L-z)^2 + \rho^2} + L - z}{\sqrt{(L+z)^2 + \rho^2} - L - z}\right)$$
where $\rho$ is defined as the distance in the picture and $R = \sqrt{\rho^2 + z^2}$

He goes on to take limits of this potential:
If z >> L then
$\phi(z,\rho) \approx \frac{\lambda}{4\pi\epsilon_0} \ln\left(\frac{1 - z/R + L/R}{1 - z/R - L/R}\right)$
and if z << L and $\rho$ << L then
$\phi(z,\rho) = \frac{\lambda}{4\pi\epsilon_0}\ln\left(\frac{\sqrt{(L)^2 + \rho^2} + L}{\sqrt{(L)^2 + \rho^2} - L}\right)$
My confusion is with the first limit. In the second limit it seems like Zangwill just said "z is small compared to L so I will drop it from the $(L - z)^2$ term and the $(L - z)$ term". But in the first limit, he only drops the L in the $(L - z)^2$ term and not the $(L-z)$ term even though z >> L. Why is this? Surely, in the limit of z >> L,  L should be dropped from both terms.
 A: I think the main reason for keeping me the $L$ even though it is sub-leading is that if you set it to zero immediately, then the numerator and denominator cancel and you get zero potential for all $z$ which is trivial.
The proper way to to take the limit is by making a Taylor expansion in $L/R$, since $R>z\gg L$.
We may then write the two fractions as 
\begin{eqnarray*}
f_{\pm} & = & \sqrt{\left(L-z\right)^{2}+\rho^{2}}-z\pm L\\
 & = & \sqrt{L^{2}-2Lz+R^{2}}-z\pm L
\end{eqnarray*}
 whre now $R,z\gg L$.
Dividing by a factor of $R$ we have 
\begin{eqnarray*}
\frac{f_{\pm}}{R} & = & \sqrt{1-\frac{2Lz}{R^{2}}+\frac{L^{2}}{R^{2}}}-\frac{z}{R}\pm\frac{L}{R}\\
 & \approx & 1-\frac{Lz}{R^{2}}+\frac{L^{2}}{2R^{2}}-\frac{z}{R}\pm\frac{L}{R}\\
 & \approx & 1-\left(\frac{L}{R}\cdot\frac{z}{R}\right)-\frac{z}{R}\pm\frac{L}{R}
\end{eqnarray*}
 where we may drop the term $\frac{L^{2}}{R^{2}}$ as it is quadratically
vanishing.
In my opition we should also keep the term $\frac{L}{R}\cdot\frac{z}{R}$ 
as it is of the same order as $\frac{L}{R}$ giving 
$$\frac{f_{\pm}}{R} = 1-\frac{z}{R}\pm\frac{L}{R}\left(1\mp\frac{z}{R}\right)
$$
instead of 
$$\frac{f_{\pm}}{R} = 1-\frac{z}{R}\pm\frac{L}{R}
$$
but there may be there considerations coming into play here.
