Theorem $:$ Suppose that $f$ is holomorphic in the region $\Omega' \subseteq \Omega$ which is obtained from $\Omega$ by a removing a single point $a \in \Omega.$ Then $f$ has a unique holomorphic extension to $\Omega$ if and only if $\lim\limits_{z \to a} (z - a) f(z) = 0.$
Uniqueness clearly follows from the fact that if the extension exists it is continuous at $a.$ It is also easy to see that the condition is necessary. In order to prove the condition is sufficient what Ahlfors did is the following $:$
Consider a circle $C$ around $a$ such that $C$ along with its interior is contained in $\Omega.$ Then by Cauchy's integral formula combined with homotopy version of Cauchy's theorem it follows that for all $z \neq a$ in the interior of $C$ $$f(z) = \frac {1} {2 \pi i} \int_{C} \frac {f(\zeta)} {\zeta - z}\ d\zeta.$$ Also since $f$ is continuous on $C,$ it turns out that the right hand side is holomorphic inside $C.$ Consequently, the function which equals to $f(z)$ for $z \neq a$ and $$\frac{1} {2 \pi i}\int_{C} \frac {f(\zeta)} {\zeta - a}\ d\zeta$$ at $z = a$ is a holomorphic extension of $f$ to $\Omega.$ $\square$
I understand the idea of the proof. The only part where I am wondering is that where did the author apply the hypothesis $\lim\limits_{z \to a} (z - a) f(z) = 0$ in the proof? I can clearly see that without that assumption the idea of extending a function holomorphically won't work. Simply consider the function $f : \mathbb C \setminus \{0\} \longrightarrow \mathbb C$ defined by $f(z) = \frac {1} {z},$ $z \in \mathbb C \setminus \{0\}.$ Then the function $f$ does not admit holomorphic extension to $\mathbb C$ simply because $\lim\limits_{z \to 0} |f(z)| = \infty.$ So the author should have used the hypothesis somewhere in the proof which I am unable to figure out. Any small suggestion regarding this would be highly solicited.
Thanks and Regards.
Ahlfors refers specifically to Theorem 3 in the words leading up to this theorem. You should pay attention to that and apply that theorem. That theorem tells us that with the given boundedness hypothesis at $z=a$ and holomorphicity elsewhere, the Cauchy theorem ($\int_{\partial R} f(z)\,dz = 0$) still holds.