A function is given as
$ln (y) = ln(\alpha)-\frac{\lambda}{\gamma}ln(\delta L^{-\gamma}+(1-\delta)K^{-\gamma})$
I need to find the second order Taylor $ln(y)$ around $\gamma=0$.
How can it be done since putting zero for gamma gives infinity?
Thank you
This is my take on a first-order approximation. If you want more, you can extend it to second order.
For small $g$ (instead of $\gamma$), $A^{-g} =e^{-g \ln A} \sim 1-g \ln A $.
Therefore, using $\ln(1-z) \sim -z$ (and $d$ instead of $\delta$),
$\begin{array}\\ \frac1{g}\ln(d L^{-g}+(1-d)K^{-g}) &\sim \frac1{g}\ln(d (1-g\ln L) + (1-d)(1-g \ln K))\\ &= \frac1{g}\ln(d -dg\ln L + (1-d)-(1-d)(g \ln K))\\ &= \frac1{g}\ln(1 -g(d\ln L +(1-d) \ln K))\\ &\sim \frac1{g}(-g(d\ln L +(1-d) \ln K)))\\ &\sim -(d\ln L +(1-d) \ln K)))\\ \end{array} $
I'll leave it at this.