Show that $\lim_{\epsilon\to0^{+}}\frac{1}{2\pi i}\int_{\gamma_\epsilon}\frac{f(z)}{z-a} \, dz=f(a)$ 
Let $a\in\Bbb C$ and $r>0$ and denote by $B(a,r)\subseteq \Bbb C$ the open ball of center $a$ and radius $r$. Assume that $f:B(a,r)\to\Bbb C$ is a continuous function and for each $\epsilon>0$ let $\gamma_\epsilon:[0,2\pi]\to \Bbb C$ be given by $\gamma_\epsilon(t)=a+\epsilon e^{it}$. Show that 
  $$\lim_{\epsilon\to0^{+}}\frac{1}{2\pi i}\int_{\gamma_\epsilon}\frac{f(z)}{z-a} \, dz = f(a).$$

I tried the following
$$\lim_{\epsilon\to0^{+}}\frac{1}{2\pi i}\int_{\gamma_\epsilon}\frac{f(z)}{z-a}dz=\lim_{\epsilon\to0^{+}}\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(a+\epsilon e^{it})}{a+\epsilon e^{it}-a}i\epsilon e^{it} \, dt = \lim_{\epsilon\to0^{+}} \frac{1}{2\pi} \int_0^{2\pi}f(a+\epsilon e^{it}) \, dt$$
If I can change the limit and the integral, then it is obvious. I tried to use the Lebesgue bounded convergence theorem to argue that, since $f:B(a,r)\to\Bbb C$ thus on $B(a,r)$ we have $|f(a+\epsilon e^{it})|\le M$, where M is the maximum of $|f(x)|$ on $B(a,r)$.
Is that valid? 

$f$ 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.
