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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 $$\lim_{\sigma \to 1} \frac{1-\sigma}{-c^{\sigma}} = \ln c?$$

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  • $\begingroup$ What is new with this function? $\endgroup$ – Guy Fsone Jan 23 '18 at 11:10
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since $$c^h =\exp(h\ln c) \sim 1+h\ln c+O(h^2)$$ Enforcing $h=1-\sigma$ gives $$\lim_{\sigma \to 1} \frac{c^{1-\sigma}-1}{1-\sigma} =\lim_{h \to0} \frac{c^{h}-1}{h}=\lim_{h \to0} \frac{h\ln c+O(h^2)}{h} =\ln c $$

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