I have given the utility function for the Ramsey-Cass-Koopmans model, as follows:

$$ u(c_t)=\frac{c^{1-\theta}-1}{1-\theta} $$

The claim is that as $\theta \rightarrow 1$, then the funciton is logarithmic, that is

$\lim_{\theta\to 1} u(c) = ln(c)$

I was told to use L'hôpital's rule, but I can't figure it out.

How can I prove this?

Thank you


2 Answers 2


The derivative of $a^{x}$ with respect to $x$ is equal to $a^x \ln(a)$.

As such, using the chain rule, the derivative of $c^{1-\theta}$ with respect to $\theta$ equals $c^{1-\theta} \ln(c) (-1)$

L'Hospital gives:

$$ \lim_{\theta \to 1} \frac{c^{1-\theta}-1}{1 - \theta}\\ = \lim_{\theta \to 1} \frac{(c^{1-\theta}-1)'}{(1-\theta)'},\\ = \lim_{\theta \to 1} \frac{c^{1-\theta} \ln(c) (-1)}{(-1)},\\ = \frac{c^0 \ln(c)}{1} = \ln(c). $$


The answer of @tdm shows you how it is done using De L'Hospital's rule. If you want to avoid this rule you can write $u(c)=\frac{c^{1-\theta}-1}{1-\theta}$ and note that for $\theta\ne 1$ this implies $u(1)=0$. Differentiating with respect to $c$ then gives you $u'(c)=c^{-\theta}$ and therefore $\lim_{\theta \to 1}u'(c)=c^{-1}$. Then integrate again to get $\lim_{\theta \to 1}u(c)=\ln(c)+K$ and comparing for $c=1$ gives you $K=0$, so $\lim_{\theta \to 1}u(c)=\ln(c)$.


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