Calculate the limit, if possible $\lim_{z \to -3i} \frac{z^3-27i}{z+3i}$
My approach was initially to calculate the conjugate and simplify, but after
$$ \frac{(z-3i)^2(z^2+3iz+9i^2)}{z^2+9}$$ it can't be simplified further.
If I continue going on, I get
$$ \frac{0}{0}$$
which seems to be wrong.
I'm suspecting this limit don't exist, any tips?
Best Answer
I recommend to use the L'Hopital rule: $$ \lim_{z \to -3i} \frac{z^3-27i}{z+3i}=\lim_{z \to -3i} \frac{(z^3-27i)^{'}}{(z+3i)^{'}}=\lim_{z \to -3i} \frac{3z^{2}}{1}=-27. $$