Show that $\lim_{\epsilon\to0^{+}}\frac{1}{2\pi i}\int_{\gamma_\epsilon}\frac{f(z)}{z-a} \, dz=f(a)$

$f$ is continuous in $a$, therefore for all $\eta > 0$ there exists a $\delta > 0$ such that $$ |f(z) - f(a)| < \eta \text{ for all } z \in B(a, \delta) \, . $$ Then for $0 < \epsilon < \delta$ $$ \left| \frac{1}{2\pi i}\int_{\gamma_\epsilon}\frac{f(z)}{z-a} \, dz - f(a) \right| = \left | \frac{1}{2\pi}\int_0^{2\pi}(f(a+\epsilon e^{it}) - f(a)) \, dt \right| \le \frac{1}{2\pi} \int_0^{2\pi} \bigl|f(a+\epsilon e^{it}) - f(a) \bigr| \, dt \\ \le \frac{1}{2\pi} \int_0^{2\pi} \eta \, dt = \eta $$ and the conclusion follows.

Hint: the function may not be bounded on $B(a,r)$ but it surely is on $\overline{B(a,r/2)}$ (which is compact), so it's not restrictive to assume it is bounded, since we're computing the limit for $\varepsilon\to0$.