I have this production function:


I am trying to find the elasticity of substitution, and I found this:

$$\sigma = -\frac{d \ln (\frac{x_2}{x_1})}{d \ln(\frac{x_1}{1+x_2})}$$

Then I have these conditions:

$x_1 >0$ and $x_2=0$

When putting this numbers in, I have ln (0) which is not possible, what is the interpretation of this elasticity of substitution?

Will the elasticity be zero or infinity?


closed as off-topic by Giskard, luchonacho, JoaoBotelho, Adam Bailey, Brythan Oct 21 '17 at 1:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not meet the standards for homework questions as spelled out in the relevant meta posts. For more information, see our policy on homework question and the general FAQ." – Giskard, luchonacho, JoaoBotelho, Adam Bailey, Brythan

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    $\begingroup$ You did not finish the derivation. Please do so, this site is not a calculator. $\endgroup$ – Giskard Oct 19 '17 at 9:02
  • $\begingroup$ Also this is not about interpretation, at best it is about limit calculus. $\endgroup$ – Giskard Oct 19 '17 at 9:03

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.

enter image description here

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.