Question: If $f(z)$ is continuous in a neighbourhood around the origin, then show that $\lim_{r\to 0}\int_{0}^{2\pi}f(re^{i\theta}) d\theta = 2\pi f(0)$.
Since the function $f$ is continuous, could I simply move the limit into the integral: $\int_{0}^{2\pi} \lim_{r\to 0}f(re^{i\theta}) d\theta = \int_{0}^{2\pi} f(0) d\theta = 2\pi f(0)$.
Is this as simple as that? or am I missing something. Thank you.
EDIT: My attempt at more rigor: Let $A = f(0)$. Since $f$ is continuous about the origin, $\forall \epsilon > 0, \exists \delta_1 >0$ such that $|z| < \delta_1 \Rightarrow |f(z) - A|< \frac{\epsilon}{2\pi}$. Since $\lim_{r\to0} re^{i\theta} = 0$, we have $\forall \epsilon> 0, \exists \delta_2$ such that $|r| < \delta_2 \Rightarrow |re^{i\theta}| < \frac{\epsilon}{2\pi}$.
Now set $\delta = min\{\delta_1, \delta_2\}$. This would mean whenever $|z = re^{i\theta}| < \delta, |f(z) - A| = |f(re^{i\theta}) - A| < \frac{\epsilon}{2\pi}$.
Now, this means if $|r|<\delta$ then $\int_0^{2\pi} |(f(re^{i\theta}) - A) d\theta| < \int_0^{2\pi} \frac{\epsilon}{2\pi} d\theta= \epsilon$.