2
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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:

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

$\endgroup$
5
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.