I have a utility function with constant elasticity of substitution (CES) that takes the form:
$$u(c)=\frac{c^{1-\sigma}-1}{1-\sigma}$$ where $\sigma >0$ is a parameter.
Why is it that when $\sigma=1$, the function becomes $u(c)=\ln(c)$? I would have thought at this value of $\sigma$ the function is undefined as subbing in the value of $\sigma=1$ results in $(1-1)/(1-1)=0/0$ which is undefined.
Any help is appreciated!
The utility function is not undefined for $\sigma=1$. You are right that in this form we cannot see what $u(c)$ is if $\sigma=1$. So we can apply the L´hôspital´s rule.
Let $f(\sigma)=c^{1-\sigma}-1$ and $g(\sigma)=1-\sigma$. Then we calculate the derivatives.
$f'(\sigma)= -c^{1 - \sigma}\cdot \log(c)$
$g'(\sigma)=-1$
$$\lim_{\sigma\to 1} \frac{f'(\sigma)}{g'(\sigma)}=\lim_{\sigma\to 1} \frac{-c^{1 - \sigma}\cdot \log(c)}{-1}=\lim_{\sigma\to 1} c^{1 - \sigma} \log(c)=\log(c)$$