Author Topic: Quiz 2 Section 6101  (Read 289 times)

Xuefen luo

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Quiz 2 Section 6101
« on: October 06, 2020, 06:50:19 AM »
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*}