I have the electric potential
$$\Phi(\vec{x})=\frac{\lambda}{2\varepsilon_0}\log\left(\frac{R+\sqrt{R^2+z^2}}{|z|}\right),$$ whose behavior I have to study in the far and near region, i.e. expand them in $z\gg R$ or $z\ll R$, respectively. The solutions provided read $$\Phi(\vec{x})=\frac{\lambda}{2\varepsilon_0}\left(\log\left(\frac{|z|}{|z|}\right)+\frac{R}{|z|}+...\right)$$ for the far region and $$\Phi(\vec{x})=\frac{\lambda}{2\varepsilon_0}\left(\log\left(\frac{2R}{|z|}\right)+...\right)$$ in the near region.
However, when I perform a Taylor expansion in these two points I do not get these results. Can anyone explain how to arrive at them?
For $|z| \ll R$ (near field), we have $$\begin{split} \frac{R+\sqrt{R^2+z^2}}{|z|} &= \frac{R+R\sqrt{1+\frac {z^2 }{R^2}}}{|z|}\\ &= \frac {R +R +\cdots}{|z|}\\ &= \frac {2R}{|z|}+\cdots \end{split}$$ This yields the answer for the near field.
For $|z| \gg R$ (far field), we have $$\begin{split} \frac{R+\sqrt{R^2+z^2}}{|z|} &= \frac{R+z\sqrt{\frac {R^2 }{z^2} +1}}{|z|}\\ &= \frac {R +z(1 + \frac {R^2}{2z^2} +\cdots)}{|z|}\\ &= \frac z {|z|} + \frac {R}{|z|} +\cdots \end{split}$$ This gives you the answer for the far field.