Limit of utility function in Ramsey-Cass-Koopmans model 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
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 $\lim_{\theta \to 1}u(c)=\ln(c)$.