$$ \ln(Y) = \ln(A) + \frac{\ln[\alpha K^\gamma + (1-\alpha) L^\gamma]}{\gamma}$$
can be taken to the limit by applying l'Hôpital's rule:
$$\lim_{\gamma\rightarrow 0} \ln(Y) = \ln(A) + \alpha \ln(K) + (1-\alpha) \ln(L).$$
I am not sure how l'Hopital's rule was used – differentiating by $\gamma$ produces some weird results.
Best Answer
Note that $$\frac{d}{d\gamma}\ln[\alpha K^\gamma + (1-\alpha) L^\gamma] = \frac{\alpha K^{\gamma}\ln (K) -(\alpha -1)L^{\gamma}\ln (L)}{\alpha K^{\gamma}-(\alpha-1)L^{\gamma}} $$ where that fact that $\frac{d}{dx} a^x = a^x \ln(a)$ is used.
Then by L'Hospital, $$\lim_{\gamma\to 0} \ln Y = \ln (A)+ \lim_{\gamma\to 0}\frac{\frac{d}{d\gamma}\ln[\alpha K^\gamma + (1-\alpha) L^\gamma]}{\frac{d}{d\gamma} \gamma}.$$