# Special Case of Neoclassical Utility Function: $\lim_{\sigma \to 1} \frac{1-\sigma}{-c^{\sigma}} = \ln c?$

The neoclassical consumption utility function is defined as

$$U(c) = \frac{c^{1-\sigma}-1}{1-\sigma}.$$

The special case of this function is $\sigma \to 1$, then the utility function converges to $$U(c) = \ln c.$$

But in order to derive this we need to solve the limit:

$$\lim_{\sigma \to 1} \frac{c^{1-\sigma}-1}{1-\sigma}.$$

I know I could start with L'Hospital rule, so I will get $$\lim_{\sigma \to 1} \frac{c^{1-\sigma}-1}{1-\sigma} = \lim_{\sigma \to 1} \frac{(1-\sigma)c^{-\sigma}}{-1} = \lim_{\sigma \to 1} \frac{1-\sigma}{-c^{\sigma}}.$$ But I have no idea how to continue. What is the next step to prove $$\lim_{\sigma \to 1} \frac{1-\sigma}{-c^{\sigma}} = \ln c?$$

• What is new with this function? – Guy Fsone Jan 23 '18 at 11:10

since $$c^h =\exp(h\ln c) \sim 1+h\ln c+O(h^2)$$ Enforcing $h=1-\sigma$ gives $$\lim_{\sigma \to 1} \frac{c^{1-\sigma}-1}{1-\sigma} =\lim_{h \to0} \frac{c^{h}-1}{h}=\lim_{h \to0} \frac{h\ln c+O(h^2)}{h} =\ln c$$