The ratio of marginal products $\frac{\partial P}{\partial x_1}/\frac{\partial P}{\partial x_2}$ (the marginal rate of tech. substitution) is equal to $1/x_1+r$ where $r\equiv x_2/x_1$.
The derivative of the MRS wrt to $r$ ($\frac{d MRS}{d r}$) is equal to $1-\frac{g}{MRS-r}$, where $g\equiv \frac{dln(x_1(r))}{dr}$.
Note how $\frac{dln(MRS)}{dr}$ which is equivalent to $\frac{d MRS/d r}{MRS}$ is equal to $\frac{MRS-r-g}{MRS(MRS-r)}$. Also note how $\frac{dln(r)}{dr}=1/r$.
Rearranging the last two relations in order to acquire expressions for $dln(MRS)$ and $dln(r)$ and substituting into the definition of the elasticity of substitution yields $-\frac{MRS(MRS-r)}{(MRS-g-r)r}$.
Finally, note how $r=0$ when $x_2=0$. Also, $MRS=1/x_1$ when $x_2=0$. The behavior of σ depends of the value of $g\equiv \frac{dln(x_1(r))}{dr}$ at $x_2=0$. If the value of $g$ is defined and is finite and constant then $σ$ approaches $- \infty$.
Intuitively, the behavior of $g$ near $x_2=0$ depends on the shape of the isocurves. $g$ denotes the (proportionate) rate of change in $x_1$ as a result of an infinitesimal change in $r$. The following graph shows that for values of $x_2$ near zero (notice the values on the vertical axis), the isocurves are really steep.

Very informally, in this setting, (really) small changes in $r$-which is equal to the ratio of $x_2/x_1$ (which is equivalent to the tangent of angle $θ$)- have almost no effect on $x_1$. For $Δθ$ approaching zero $Δx_1$ is also small and probably close to zero too. Equivalently and very loosely speaking, in this region $g$ should be close to zero too.