"By hypothesis, for any $0<\epsilon<a$ there exists $\delta(\epsilon)>0$ such that for any $x \in \text{dom}(f) \cap (x_0-\delta(\epsilon),x_0+\delta(\epsilon)) \setminus \{x_0\}$, $|x-x_0|<\delta(\epsilon) \implies |f(x)-l|<\epsilon$.

Suppose $\epsilon' \ge a$. Using the limit hypothesis with $0<\frac{a-\epsilon}{2}<a$, there exists $\delta'=\delta((a-\epsilon)/2)>0$ such that for any $x \in \text{dom}(f) \cap (x_0-\delta',x_0+\delta') \setminus \{x_0\}$, $|x-x_0|<\delta' \implies |f(x)-l|<(a-\epsilon)/2<a\le\epsilon'$. Since $\epsilon' \ge a$ is arbitrary, this holds for any $\epsilon' \…

In more detail, just think of the complement of the link $L$ inside $S^1 \times D^2$. What you drew here then is a literal surface $S$ bounding $\{\theta\} \times \partial D^2$ in this link complement. Once you do a $0$-surgery on the two link components, you can "push" $S$ inside one of the added torii post-surgery -- which one will it go inside, as there are two? Well, it's the one which has both the circles $(\{\theta\} \times D^2) \cap \partial\nu(L)$ as a meridian (assuming $\theta$ is one for which $(\{\theta\} \times D^2) \cap L$ is $2$ points), where $\nu(L)$ is a thickening of $L$ …