Find the limit of the function at the given point, or explain why it does not exist.
\begin{align*}
f(z)&=\frac{z^3-8i}{z+2i} \ (z \neq -2i) \ at \ z_0=-2i\\
\end{align*}
Answer:
\begin{align*}
\lim_{z \to -2i} f(z) &= \lim_{z \to -2i} \frac{z^3-8i}{z+2i}\\
&=\lim_{z \to -2i} \frac{z^3+(2i)^3}{z+2i}\\
&=\lim_{z \to -2i} \frac{(z+2i)(z^2-2iz-4)}{z+2i} \ (By\ a^3+b^3=(a+b)(a^2-ab+b^2))\\
&=\lim_{z \to -2i} (z^2-2iz-4)\\
&= (-2i)^2-2i(-2i)-4\\
&= -4-4-4\\
&= -12\\
\end{align*}
