Why do we use this particular approximation for the bilinear transform?

As I understand it, for a signal $f(t)$ in time, its Laplace transform $\mathfrak{L}\left\{f(t)\right\}=F_1(s)$ and Z transform $\mathfrak{Z}\left\{f(t)\right\}=F_2(z)$ are related by a transformation $z=e^{sT}\leftrightarrow s=1/T\,\log(z)$ where $T$ is the sampling period (since the Z transform is discrete in time).

In practice, this is approximated to the first degree as follows $$\begin{align*}z&=e^{sT}\\&=\frac{e^{sT/2}}{e^{-sT/2}}\\&\approx\frac{1+sT/2}{1-sT/2}\end{align*}$$ and thus $(1-sT/2)z\approx1+sT/2$ so $sT/2\approx(z-1)/(z+1)$ and ultimately $s\approx\frac2T\frac{z-1}{z+1}=\frac2T\frac{1-z^{-1}}{1+z^{-1}}$.

Now, I understand up to here, but I fail to understand why we use this particular first-order approximation over, say, $z=e^{sT}\approx1+sT\leftrightarrow s\approx(z-1)/T=\frac{1-z^{-1}}{Tz^{-1}}$.

Does this approximation 'behave' in some significantly poorer way for most purposes?

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Sorry about the tags -- I tried various things like 'bilinear-transform' but they did not exist and I lack the points to create them.