# RBC model and indivisible labour

First of all, I would like to apologise for the fact that I will include a link in my question rather than a direct question, however, I have no idea how to do it in a different way.

I have been reading this paper recently: RBC extensions ,

I don't understand one thing completely. On the second page when we are reducing the equation for utility, why cannot we just drop the term: $$\theta \frac{(1)^{1-\xi}-1}{1-\xi}$$ I was wondering how this could possibly work, because common-sensically: $$(1)^{1-\xi}=1 \Rightarrow \theta \frac{(1)^{1-\xi}-1}{1-\xi}=0$$ Am I wrong?

Thank you guys!

The only reason I can see right now could possibly be pedagogical. While that term does indeed just equal zero, there could be generalizations where the agent works a high amount with probability $\tau$ and a low, but positive, amount with probability $1-\tau$. In this case, the term would still go through. There are also plenty of functional forms that would result in that term not being zero (taking away the -1, for example, which is often done in applied work). At any rate, this is an idiosyncratic enough case that the professor might have thought it best to show the process of rearranging terms for future reference and to give help with other scenarios. On the other hand, he may just not have noticed it. :)
• Thank you for your answer. I would also like to ask you another question if I can. The professor is also writing that when $\xi=1$, then $\theta \frac{(1-n_t)^{1-\xi}-1}{1-\xi}$ becomes $log(1-n_t)$ (top of the second page). What is the maths behind this? It is quite apparent that when $\xi=1$, then $\theta \frac{(1-n_t)^{1-\xi}-1}{1-\xi}=\frac{0}{0}$. I cannot get this. – marco11 Jun 14 '15 at 11:27
• That's actually a very common result that can be proved by taking the limit of the function as $\xi$ goes to 1. I would suggest asking this in a new question so the proof can be given the appropriate treatment. – philE Jun 15 '15 at 16:48